Online Self Test
Learn & Practice Quizzes & Challenges Competitive Exams
 

Current Electricity – CBSE NCERT Study Resources

Previous ChapterNext Chapter
Previous Chapter Next Chapter

Study Materials

12th

12th - Physics

Current Electricity

Jump to Question Sets

Overview of the Chapter: Current Electricity

This chapter introduces the fundamental concepts of electric current, its behavior in conductors, and the principles governing electrical circuits. Students will explore Ohm's Law, resistivity, Kirchhoff's laws, and the working of electrical devices like potentiometers and meters.

Electric Current: The flow of electric charge through a conductor per unit time. It is measured in amperes (A).

Key Topics Covered

  • Electric Current and Drift Velocity
  • Ohm's Law and Resistance
  • Electrical Energy and Power
  • Combination of Resistors (Series and Parallel)
  • Kirchhoff's Laws and Their Applications
  • Potentiometer and Its Uses

Detailed Concepts

Electric Current and Drift Velocity

The flow of electric charges constitutes an electric current. The average velocity of charge carriers in a conductor under an electric field is called drift velocity.

Drift Velocity (vd): The average velocity attained by charged particles due to an electric field, given by vd = (eEτ)/m, where e is charge, E is electric field, τ is relaxation time, and m is mass.

Ohm's Law and Resistance

Ohm's Law states that the current (I) through a conductor is directly proportional to the potential difference (V) across it, provided physical conditions remain constant.

Resistance (R): The opposition offered by a conductor to the flow of current, given by R = V/I. Its unit is ohm (Ω).

Electrical Energy and Power

The rate at which electrical energy is consumed in a circuit is called electric power (P = VI). Energy dissipated is given by H = I²Rt (Joule's Law).

Combination of Resistors

Resistors can be combined in series or parallel:

  • Series: Req = R1 + R2 + ... + Rn
  • Parallel: 1/Req = 1/R1 + 1/R2 + ... + 1/Rn

Kirchhoff's Laws

These laws help analyze complex circuits:

Kirchhoff's Current Law (KCL): The algebraic sum of currents at any junction is zero.

Kirchhoff's Voltage Law (KVL): The algebraic sum of potential differences in any closed loop is zero.

Potentiometer

A device used to measure emf of a cell or compare emfs of two cells without drawing current from them.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
Define electric current.
Answer:

Flow of electric charge per unit time.

Question 2:
What is the condition for balanced Wheatstone bridge?
Answer:

P/Q = R/S.

Question 3:
Define internal resistance of a cell.
Answer:

Opposition to current flow within the cell itself.

Question 4:
State Kirchhoff's junction rule.
Answer:

Sum of currents entering a junction equals sum leaving.

Question 5:
What is the purpose of potentiometer?
Answer:

To measure unknown emf without drawing current.

Question 6:
State Ohm's Law.
Answer:

Current is directly proportional to voltage at constant temperature.

Question 7:
What is the SI unit of resistivity?
Answer:

Ohm-meter (Ω·m).

Question 8:
Name the device used to measure potential difference.
Answer:

Voltmeter.

Question 9:
What happens to resistance when wire length is doubled?
Answer:

Resistance doubles.

Question 10:
Define drift velocity of electrons.
Answer:

Average velocity of electrons in a conductor under electric field.

Question 11:
Give one example of a non-ohmic conductor.
Answer:

Semiconductor diode.

Question 12:
What is the effect of temperature on metal resistance?
Answer:

Resistance increases with temperature.

Question 13:
Write the formula for electrical power.
Answer:

P = VI.

Question 14:
Name the material with negative temperature coefficient of resistance.
Answer:

Semiconductor.

Question 15:
Define electric current and state its SI unit.
Answer:

The electric current is the rate of flow of electric charge through a conductor.
Its SI unit is the ampere (A).

Question 16:
What is the direction of conventional current in a circuit?
Answer:

The conventional current flows from the positive terminal to the negative terminal of the battery, opposite to the flow of electrons.

Question 17:
State Ohm's Law and write its mathematical expression.
Answer:

Ohm's Law states that the current (I) flowing through a conductor is directly proportional to the potential difference (V) across it, provided the physical conditions remain constant.
Mathematically: V = IR, where R is the resistance.

Question 18:
What is the resistance of an ideal ammeter?
Answer:

An ideal ammeter has zero resistance to ensure it does not alter the current in the circuit.

Question 19:
Name the material used for making the filament of an electric bulb.
Answer:

The filament of an electric bulb is made of tungsten due to its high melting point and resistivity.

Question 20:
What happens to the resistance of a conductor if its length is doubled?
Answer:

The resistance of a conductor doubles when its length is doubled, as resistance is directly proportional to length (R ∝ l).

Question 21:
Define resistivity and state its SI unit.
Answer:

Resistivity is the inherent property of a material that quantifies its resistance per unit length and cross-sectional area.
Its SI unit is ohm-meter (Ω·m).

Question 22:
Why are alloys commonly used in heating devices?
Answer:

Alloys are used in heating devices because they have high resistivity and low temperature coefficient of resistance, ensuring stable performance at high temperatures.

Question 23:
What is the purpose of using a fuse in an electric circuit?
Answer:

A fuse protects the circuit by melting and breaking the circuit when the current exceeds a safe limit, preventing damage to appliances.

Question 24:
How does the resistance of a conductor vary with temperature?
Answer:

The resistance of a conductor increases with temperature due to increased vibrations of atoms, which hinder the flow of electrons.

Question 25:
What is the equivalent resistance of two resistors connected in series?
Answer:

The equivalent resistance (Req) of two resistors R1 and R2 in series is:
Req = R1 + R2.

Question 26:
Why is copper preferred over aluminum for electrical wiring?
Answer:

Copper is preferred because it has lower resistivity and higher ductility, making it more efficient and easier to use in wiring.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
What is the difference between drift velocity and thermal velocity of electrons?
Answer:

Drift velocity is the average velocity of electrons in a conductor due to an applied electric field, typically very small (~10-4 m/s).
Thermal velocity is the random motion of electrons due to thermal energy, much higher (~105 m/s) but averages to zero.

Question 2:
State Ohm's Law and write its mathematical expression.
Answer:

Ohm's Law states that the current (I) flowing through a conductor is directly proportional to the potential difference (V) across it, provided physical conditions remain constant.
Mathematically: V = IR, where R is resistance.

Question 3:
Why does the resistance of a conductor increase with temperature?
Answer:

With increasing temperature, the thermal vibrations of the conductor's lattice ions increase, causing more frequent collisions with free electrons.
This scattering effect reduces electron mobility, increasing resistance.

Question 4:
What is the significance of the temperature coefficient of resistance?
Answer:

The temperature coefficient of resistance (α) quantifies how much a material's resistance changes per degree temperature change.
For metals, α is positive (resistance increases with temperature), while for semiconductors, it is negative.

Question 5:
Explain why a potentiometer is considered more accurate than a voltmeter for measuring emf.
Answer:

A potentiometer measures emf by balancing it against a known potential difference, drawing no current from the cell at null point.
Voltmeters draw current, causing internal resistance errors, making potentiometers more precise.

Question 6:
Define Kirchhoff's junction rule and state its physical basis.
Answer:

Kirchhoff's junction rule states that the algebraic sum of currents at any junction is zero (ΣI = 0).
It is based on the conservation of charge, as charge cannot accumulate or vanish at a point.

Question 7:
What is the Wheatstone bridge principle? Give its balancing condition.
Answer:

The Wheatstone bridge measures unknown resistance by balancing two legs of a circuit.
At balance: P/Q = R/S, where P, Q are known resistances and R is unknown, S is variable.

Question 8:
Why are alloys like constantan or manganin used for standard resistors?
Answer:

Alloys like constantan or manganin have very low temperature coefficients and high resistivity.
This ensures minimal resistance variation with temperature, making them ideal for precise resistors.

Question 9:
How does the resistivity of a semiconductor change with doping?
Answer:

Doping a semiconductor introduces impurities that increase charge carriers (electrons/holes).
Higher carrier concentration reduces resistivity, making the material more conductive.

Question 10:
Distinguish between emf and terminal voltage of a cell.
Answer:

Emf is the maximum potential difference when no current flows (energy supplied per unit charge).
Terminal voltage is the actual voltage across cell terminals during current flow, reduced by internal resistance (V = E - Ir).

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Define drift velocity of electrons in a conductor. How does it relate to the electric current flowing through the conductor?
Answer:

The drift velocity is the average velocity acquired by free electrons in a conductor when an electric field is applied. It is denoted by vd and is given by the formula: vd = (eEτ)/m, where e is the charge of an electron, E is the electric field, τ is the relaxation time, and m is the mass of the electron.

The relationship between drift velocity and electric current (I) is given by: I = nAevd, where n is the number density of electrons, A is the cross-sectional area of the conductor, and e is the electronic charge. This shows that current is directly proportional to drift velocity.

Question 2:
Explain why the resistance of a conductor increases with temperature, while that of a semiconductor decreases.
Answer:

In a conductor, resistance increases with temperature because:
1. As temperature rises, the lattice vibrations of the conductor's atoms increase.
2. This causes more frequent collisions between free electrons and the lattice ions.
3. The mean free path of electrons decreases, leading to higher resistance.

In a semiconductor, resistance decreases with temperature because:
1. Higher temperature provides more energy to valence electrons, breaking covalent bonds.
2. This increases the number of charge carriers (electrons and holes).
3. The rise in charge carriers outweighs the effect of increased lattice vibrations, reducing resistance.

Question 3:
State Ohm's Law. Under what conditions is it valid?
Answer:

Ohm's Law states that the current (I) flowing through a conductor is directly proportional to the potential difference (V) applied across it, provided the physical conditions (like temperature) remain constant. Mathematically, V = IR, where R is the resistance.

Conditions for validity:
1. The conductor must be at a constant temperature.
2. The material must obey linear V-I characteristics (e.g., metallic conductors).
3. It does not apply to non-ohmic materials like diodes or semiconductors under varying conditions.

Question 4:
Derive the expression for the equivalent resistance of three resistors connected in parallel.
Answer:

For three resistors R1, R2, and R3 connected in parallel:
1. The potential difference (V) across each resistor is the same.
2. Total current I = I1 + I2 + I3 (from Kirchhoff's current law).
3. Using Ohm's Law: I1 = V/R1, I2 = V/R2, I3 = V/R3.
4. Substituting: I = V(1/R1 + 1/R2 + 1/R3).
5. Equivalent resistance Req is given by V = IReq.
6. Thus, 1/Req = 1/R1 + 1/R2 + 1/R3.

Question 5:
What is the significance of the temperature coefficient of resistance? Write its formula and unit.
Answer:

The temperature coefficient of resistance (α) measures how much a material's resistance changes with temperature. It indicates whether resistance increases or decreases with temperature and by how much.

Formula: α = (Rt - R0)/(R0 × ΔT), where:
- Rt = resistance at temperature t,
- R0 = resistance at reference temperature (usually 0°C or 20°C),
- ΔT = change in temperature.

Unit: °C-1 or K-1 (per degree Celsius or Kelvin).

Question 6:
Describe how a potentiometer is used to compare the emfs of two primary cells.
Answer:

To compare emfs (E1 and E2) of two cells using a potentiometer:
1. Connect the cells individually to the potentiometer circuit.
2. Adjust the sliding contact to find the null point (where galvanometer shows zero deflection).
3. Measure the balancing lengths L1 and L2 for each cell.
4. Since E ∝ L at null point, the ratio of emfs is: E1/E2 = L1/L2.

Advantages:
- No current is drawn from the cell at null point, ensuring accurate measurement.
- Internal resistance of the cell does not affect the comparison.

Question 7:
Define drift velocity of electrons in a conductor and derive its relation with electric current.
Answer:

The drift velocity is the average velocity acquired by free electrons in a conductor under the influence of an electric field.


Relation with current (I):


I = nAevd

Where:
n = number density of electrons
A = cross-sectional area
e = charge of electron
vd = drift velocity


Derivation:
Total charge flowing through conductor in time Δt: ΔQ = (nA vd Δt) e
Current I = ΔQ/Δt = nAevd

Question 8:
Explain how the resistance of a conductor varies with temperature using the concept of resistivity.
Answer:

The resistivity (ρ) of a conductor increases with temperature due to increased lattice vibrations.


Mathematical relation:
ρT = ρ0 [1 + α(T - T0)]
Where:
α = temperature coefficient of resistivity
T = final temperature
T0 = reference temperature


For conductors:

  • Increased temperature → more electron-lattice collisions
  • Mean free path decreases → resistivity increases
  • Resistance R = ρL/A thus increases

Question 9:
State Kirchhoff's laws and explain their significance in analyzing electrical circuits.
Answer:

Kirchhoff's Current Law (KCL): The algebraic sum of currents meeting at any junction is zero.
Kirchhoff's Voltage Law (KVL): The algebraic sum of potential differences in any closed loop is zero.


Significance:

  • KCL conserves charge at junctions
  • KVL conserves energy in closed loops
  • Essential for analyzing complex circuits with multiple components
  • Forms basis for network theorems like Thevenin's theorem
  • Used in designing and troubleshooting electrical circuits

Question 10:
Derive the expression for equivalent resistance of three resistors connected in parallel.
Answer:

For resistors R1, R2, and R3 in parallel:


Voltage across each resistor is same (V)
Total current I = I1 + I2 + I3
Using Ohm's Law:
V/Req = V/R1 + V/R2 + V/R3
Canceling V:
1/Req = 1/R1 + 1/R2 + 1/R3
Thus:
Req = (R1R2R3)/(R1R2 + R2R3 + R3R1)

Question 11:
Describe the principle and working of a potentiometer to compare EMFs of two cells.
Answer:

Principle: A potentiometer measures potential difference by balancing it against a known voltage drop across a uniform wire.


Procedure to compare EMFs:

  • Connect the first cell (E1) and find balancing length L1
  • Connect second cell (E2) and find balancing length L2
  • Since E ∝ L, the ratio is E1/E2 = L1/L2

Advantages over voltmeter:

  • No current drawn from cell during measurement
  • More accurate as it uses null method
  • Can measure small potential differences precisely

Question 12:
Explain how the internal resistance of a cell affects its terminal voltage when connected to a load.
Answer:

When a cell with EMF E and internal resistance r is connected to load R:


Terminal voltage V = E - Ir
Where I = E/(R + r) is the circuit current


Effects:

  • Higher r causes greater voltage drop (Ir)
  • Terminal voltage decreases as current increases
  • Maximum current is limited by Imax = E/r (when R = 0)

Practical implications:

  • Cells heat up due to energy loss in r
  • Efficiency decreases with higher r
  • Voltage regulation becomes poorer

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Explain the Kirchhoff’s laws and derive the condition for balanced Wheatstone bridge using these laws.
Answer:
Theoretical Framework

We studied Kirchhoff’s laws, which include the junction rule (current conservation) and the loop rule (energy conservation). These laws help analyze complex circuits.

Evidence Analysis
  • For a balanced Wheatstone bridge, no current flows through the galvanometer, implying equal potential at its ends.
  • Applying Kirchhoff’s loop rule to two loops, we derive the condition: P/Q = R/S, where P, Q, R, S are resistances.
Critical Evaluation

This derivation assumes ideal conditions, but real-world factors like wire resistance may affect accuracy.

Future Implications

Wheatstone bridges are used in strain gauges and sensor calibrations, highlighting their practical relevance.

Question 2:
Define drift velocity and derive its relation with current density. Discuss how temperature affects it.
Answer:
Theoretical Framework

Drift velocity (vd) is the average velocity of electrons in a conductor under an electric field. Current density (J) is current per unit area.

Evidence Analysis
  • We derived J = nevd, where n is electron density and e is charge.
  • As temperature rises, electron collisions increase, reducing vd and increasing resistivity.
Critical Evaluation

This model simplifies electron behavior, ignoring quantum effects at low temperatures.

Future Implications

Understanding drift velocity aids in designing efficient conductors for high-temperature applications.

Question 3:
Compare series and parallel combinations of resistors. Which is preferable for household wiring and why?
Answer:
Theoretical Framework

In series, resistors share the same current, while in parallel, they share the same voltage. Total resistance differs in each case.

Evidence Analysis
  • Series: Rtotal = R1 + R2 + ...; Parallel: 1/Rtotal = 1/R1 + 1/R2 + ....
  • Parallel is preferred for homes as it ensures independent operation of appliances and uniform voltage.
Critical Evaluation

Parallel circuits reduce overall resistance, increasing current load, which requires proper circuit protection.

Future Implications

Smart grids use parallel configurations to optimize energy distribution in modern homes.

Question 4:
Describe the Ohm’s law and its limitations. Give two examples where it fails.
Answer:
Theoretical Framework

Ohm’s law states V = IR, where V is voltage, I is current, and R is resistance, assuming constant temperature.

Evidence Analysis
  • Limitations: Non-ohmic materials (e.g., diodes) show nonlinear V-I curves.
  • Examples: Semiconductors and electrolytes deviate due to temperature and ion mobility.
Critical Evaluation

Ohm’s law is foundational but oversimplifies real-world material behavior.

Future Implications

Advanced electronics rely on non-ohmic components, necessitating broader models like Shockley’s diode equation.

Question 5:
Explain the principle of potentiometer and how it measures internal resistance of a cell.
Answer:
Theoretical Framework

A potentiometer balances unknown EMF against a known voltage drop across a uniform wire, ensuring no current flow during measurement.

Evidence Analysis
  • To find internal resistance (r), we first balance the cell’s EMF, then connect a known resistor and rebalance.
  • Using r = R(l1 - l2)/l2, where R is external resistance and l1, l2 are balancing lengths.
Critical Evaluation

This method assumes wire uniformity, which may degrade over time.

Future Implications

Potentiometers are used in precision instruments, emphasizing their accuracy in low-current measurements.

Question 6:
Explain Ohm's Law and derive its mathematical expression. Discuss its limitations and practical applications.
Answer:
Theoretical Framework

Ohm's Law states that the current (I) through a conductor is directly proportional to the potential difference (V) across it, provided physical conditions remain constant. Mathematically, V = IR, where R is resistance.

Evidence Analysis
  • Our textbook shows metallic conductors obey Ohm's Law at constant temperature.
  • Non-ohmic devices like diodes violate this law.
Critical Evaluation

Limitations include temperature dependence and inapplicability to semiconductors. Practical uses include resistor networks and circuit design.

Future Implications

Advanced materials research may develop new ohmic components for nanoelectronics.

Question 7:
Compare series and parallel combinations of resistors with circuit diagrams. Which arrangement is used in household wiring and why?
Answer:
Theoretical Framework

In series, resistors share current (Req = R1+R2). In parallel, they share voltage (1/Req = 1/R1+1/R2).

[Diagram: Series vs parallel circuits]Evidence Analysis
  • Series circuits dim when adding bulbs (increased R)
  • Parallel maintains brightness (independent paths)
Critical Evaluation

Homes use parallel: independent appliance operation and consistent voltage supply.

Future Implications

Smart grids optimize parallel configurations for energy efficiency.

Question 8:
Derive the expression for equivalent resistance of three resistors in parallel. How does this affect total current in the circuit?
Answer:
Theoretical Framework

For resistors R1, R2, R3 in parallel: 1/Req = 1/R1 + 1/R2 + 1/R3. Current divides inversely with resistance.

Evidence Analysis
  • Lab experiments show Req < smallest individual R
  • Our textbook demonstrates current conservation: Itotal = I1+I2+I3
Critical Evaluation

Parallel grouping reduces effective resistance, increasing total current per Ohm's Law.

Future Implications

Power distribution systems leverage this principle for load balancing.

Question 9:
Describe the Wheatstone bridge principle with circuit diagram. Explain its use in measuring unknown resistance.
Answer:
Theoretical Framework

Wheatstone bridge balances when P/Q = R/S (null deflection in galvanometer). [Diagram: Bridge circuit]

Evidence Analysis
  • Our textbook shows precision resistance measurement to ±0.1%
  • Practical applications include strain gauges and thermistors
Critical Evaluation

Advantages include high accuracy; limitations require known resistances and sensitive galvanometer.

Future Implications

Modern digital multimeters adapt this principle for automated measurements.

Question 10:
Explain Kirchhoff's laws with examples. How do they help analyze complex circuits?
Answer:
Theoretical Framework

Kirchhoff's Current Law (KCL): ΣIin = ΣIout at junctions. Voltage Law (KVL): ΣV = 0 in closed loops.

Evidence Analysis
  • Example 1: Solving for currents in multi-battery circuits
  • Example 2: Analyzing bridge networks beyond balance condition
Critical Evaluation

These laws enable systematic analysis where Ohm's Law alone fails, especially in non-serial networks.

Future Implications

Circuit simulation software implements these laws for IC design verification.

Question 11:
Discuss drift velocity of electrons in conductors. How does it relate to current density and electric field?
Answer:
Theoretical Framework

Drift velocity (vd) is net electron motion under electric field: vd = (eEτ)/m where τ is relaxation time.

Evidence Analysis
  • For copper, vd ≈ 0.1mm/s at 1A/mm2
  • Current density J = nevd (n=charge density)
Critical Evaluation

Despite slow vd, instantaneous field propagation enables fast current establishment.

Future Implications

Nanowire research examines drift velocity limits for high-speed electronics.

Question 12:
Explain the principle of a potentiometer and derive the expression for the comparison of emfs of two cells using a potentiometer. Highlight the advantages of a potentiometer over a voltmeter.
Answer:

A potentiometer is a device used to measure the emf (electromotive force) of a cell or to compare the emfs of two cells without drawing any current from them. It works on the principle that the potential difference across a uniform wire is directly proportional to its length when a constant current flows through it.


Derivation:
Let the emfs of the two cells be E1 and E2. When the first cell is connected, the balancing length is l1, and for the second cell, it is l2.
For the first cell: E1 = k l1
For the second cell: E2 = k l2
Dividing the two equations: E1/E2 = l1/l2
Thus, the emfs can be compared by measuring the balancing lengths.


Advantages of a potentiometer over a voltmeter:

  • It measures the emf of a cell accurately without drawing any current, whereas a voltmeter draws some current, leading to a slight error.
  • It can be used to compare emfs of two cells directly, which a voltmeter cannot do.
  • The potentiometer is more sensitive and can detect small potential differences.

Question 13:
Define Kirchhoff's laws and apply them to calculate the current in each branch of the given circuit. Also, explain the significance of these laws in analyzing complex circuits.
Answer:

Kirchhoff's laws are fundamental for analyzing electrical circuits. They consist of two rules:
1. Kirchhoff's Current Law (KCL): The algebraic sum of currents at any junction is zero.
2. Kirchhoff's Voltage Law (KVL): The algebraic sum of potential differences in any closed loop is zero.


Application to a given circuit:
Consider a circuit with two resistors R1 and R2 in parallel, connected to a battery of emf E.
Let the currents through R1 and R2 be I1 and I2, respectively.
Using KCL at the junction: I = I1 + I2
Using KVL in the loop containing R1: E - I1R1 = 0
Using KVL in the loop containing R2: E - I2R2 = 0
Solving these equations gives: I1 = E/R1 and I2 = E/R2.


Significance of Kirchhoff's laws:

  • They provide a systematic method to analyze complex circuits with multiple loops and junctions.
  • They are based on the conservation of charge (KCL) and energy (KVL), making them universally applicable.
  • They help in deriving circuit equations that can be solved to find unknown currents or voltages.

Question 14:
Explain the concept of drift velocity of electrons in a conductor. Derive the expression for current in terms of drift velocity and discuss its significance in the context of Ohm's Law.
Answer:

The drift velocity is the average velocity acquired by free electrons in a conductor when an external electric field is applied. Due to collisions with ions in the conductor, electrons do not accelerate indefinitely but attain a steady average velocity called drift velocity (vd).


Derivation of current in terms of drift velocity:


1. Consider a conductor of length l, cross-sectional area A, and number density of free electrons n.
2. Total number of free electrons in the conductor = nAl.
3. If e is the charge of an electron, total charge (Q) = nAle.
4. Time taken for electrons to drift across the conductor = t = l/vd.
5. Current (I) = Q/t = nAle/(l/vd) = nAevd.


Significance in Ohm's Law: The drift velocity is directly proportional to the electric field (E), i.e., vd ∝ E. Since current I = nAevd, it implies I ∝ E, which is the essence of Ohm's Law (V = IR). This relationship helps explain why conductors obey Ohm's Law under constant temperature conditions.

Question 15:
Describe the working principle of a potentiometer. How is it used to compare the emfs of two primary cells? Derive the necessary formula and explain why a potentiometer is considered more accurate than a voltmeter for such measurements.
Answer:

The potentiometer works on the principle that the potential difference across a uniform wire is directly proportional to its length when a constant current flows through it. It is used to measure unknown emfs by balancing them against a known potential difference.


Procedure to compare emfs of two primary cells (E1 and E2):

  • Connect the two cells in series with a galvanometer and a jockey on the potentiometer wire.
  • Adjust the jockey to find the null point (where galvanometer shows zero deflection) for each cell.
  • Let the balancing lengths be l1 (for E1) and l2 (for E2).

Derivation of formula: At null point, E1 = k l1 and E2 = k l2, where k is the potential gradient.
Thus, E1/E2 = l1/l2.


Advantage over voltmeter: A potentiometer does not draw any current from the cell at the null point, ensuring accurate measurement of emf. In contrast, a voltmeter draws a small current, leading to potential drop across internal resistance and inaccurate readings.

Question 16:
Explain the principle of a potentiometer and derive the expression for the comparison of emf of two cells using a potentiometer. Discuss its advantages over a voltmeter.
Answer:

A potentiometer is a device used to measure the emf (electromotive force) of a cell accurately by balancing it against a known potential difference. It works on the principle that the potential drop across a uniform wire is directly proportional to its length when a constant current flows through it.


Derivation for comparing emf of two cells:


Let E1 and E2 be the emfs of two cells to be compared. The potentiometer wire AB is of length L and uniform cross-section. A constant current is passed through it.


1. First, connect cell E1 to the potentiometer. Adjust the jockey to find the null point at length l1.
At null point, E1 = k l1, where k is the potential gradient.


2. Now, replace E1 with E2 and find the new null point at length l2.
At null point, E2 = k l2.


3. Dividing the two equations:
E1/E2 = l1/l2.


Advantages over a voltmeter:

  • Potentiometer measures emf directly without drawing current from the cell, while a voltmeter draws some current, leading to inaccurate readings.
  • It provides high accuracy as it uses a null method (no current flows at the balance point).
  • It can measure small potential differences effectively.
Question 17:
Define Ohm's Law and derive the expression for the equivalent resistance of three resistors connected in parallel. Also, state the limitations of Ohm's Law.
Answer:

Ohm's Law states that the current (I) flowing through a conductor is directly proportional to the potential difference (V) across its ends, provided the physical conditions (like temperature) remain constant. Mathematically, V = IR, where R is the resistance of the conductor.


Derivation for equivalent resistance in parallel:


Consider three resistors R1, R2, and R3 connected in parallel across a potential difference V.


1. The total current I divides into three branches: I = I1 + I2 + I3.
Using Ohm's Law, I1 = V/R1, I2 = V/R2, and I3 = V/R3.


2. Substituting these into the current equation:
I = V/R1 + V/R2 + V/R3.


3. Let the equivalent resistance be Req. Then, I = V/Req.
Thus, V/Req = V(1/R1 + 1/R2 + 1/R3).


4. Canceling V from both sides:
1/Req = 1/R1 + 1/R2 + 1/R3.


Limitations of Ohm's Law:

  • It is not applicable to non-ohmic conductors like diodes and transistors, where the V-I relationship is nonlinear.
  • It fails at very high temperatures or when the physical conditions of the conductor change.
  • It does not apply to semiconductors or insulators.
Question 18:
Explain the principle of a potentiometer and derive the expression for the comparison of emf of two primary cells using it. Discuss why a potentiometer is considered more accurate than a voltmeter for measuring emf.
Answer:

A potentiometer is a device used to measure the emf (electromotive force) of a cell or compare the emf of two cells without drawing any current from them. It works on the principle that the potential difference across a uniform wire is directly proportional to its length when a constant current flows through it.


Principle: When a steady current flows through a uniform wire of length L, the potential drop (V) across any segment of the wire is proportional to the length (l) of that segment, i.e., V ∝ l.


Derivation for comparing emf of two cells:
1. Connect the two primary cells (E1 and E2) to the potentiometer circuit alternately.
2. Adjust the sliding contact to find the null point (where galvanometer shows zero deflection) for each cell.
3. Let the balancing lengths be l1 for E1 and l2 for E2.
4. Since the potential gradient (k) is constant, E1 = k l1 and E2 = k l2.
5. Dividing the two equations, we get: E1/E2 = l1/l2.


Why potentiometer is more accurate than a voltmeter:
1. A potentiometer measures emf under null condition, where no current is drawn from the cell, ensuring no internal resistance error.
2. A voltmeter draws some current, leading to potential drop across the cell's internal resistance, giving a lower reading than the actual emf.
3. The potentiometer's accuracy depends only on the uniformity of the wire and the calibration of the standard cell, making it highly precise.

Question 19:
Explain the principle of a potentiometer and derive the expression for the comparison of emfs of two cells using a potentiometer. Also, state two advantages of a potentiometer over a voltmeter.
Answer:

A potentiometer is a device used to measure the emf (electromotive force) of a cell or to compare the emfs of two cells accurately without drawing any current from them. It works on the principle that the potential difference across a uniform wire is directly proportional to its length when a constant current flows through it.


Derivation for comparison of emfs:


1. Let E1 and E2 be the emfs of the two cells to be compared.
2. Connect the first cell (E1) to the potentiometer and find the balancing length l1 where the galvanometer shows null deflection.
3. The potential gradient along the wire is given by k = V/L, where V is the potential difference across the wire and L is its total length.
4. At balance, E1 = k l1.
5. Repeat the process for the second cell (E2) to find the balancing length l2, giving E2 = k l2.
6. Dividing the two equations, we get: E1/E2 = l1/l2.


Advantages of a potentiometer over a voltmeter:

  • It measures the emf of a cell without drawing any current from it, ensuring accurate results.
  • It can measure very small potential differences with high precision, unlike a voltmeter which has limited sensitivity.
Question 20:
Explain the principle of a potentiometer and derive the expression for the comparison of emfs of two cells using a potentiometer. Also, state the advantages of using a potentiometer over a voltmeter for measuring emf.
Answer:

A potentiometer is a device used to measure the emf (electromotive force) of a cell or to compare the emfs of two cells accurately. It works on the principle that the potential difference across a length of a uniform wire carrying a constant current is directly proportional to its length.

Principle: When a steady current flows through a uniform wire of high resistance, the potential drop across any segment of the wire is directly proportional to the length of that segment.

Derivation for comparing emfs of two cells:
1. Let E1 and E2 be the emfs of the two cells to be compared.
2. Connect the cells in series with a galvanometer and a jockey on the potentiometer wire.
3. Adjust the rheostat to get a constant current in the main circuit.
4. Find the balancing lengths l1 and l2 for cells E1 and E2 respectively, where the galvanometer shows null deflection.
5. Since the potential gradient is constant, E1 ∝ l1 and E2 ∝ l2.
6. Therefore, the ratio of emfs is given by: E1/E2 = l1/l2.

Advantages of potentiometer over voltmeter:

  • A potentiometer measures the true emf of a cell as it draws no current from the cell at the null point.
  • Voltmeters draw some current, leading to potential drop across internal resistance, resulting in inaccurate readings.
  • Potentiometers provide higher precision and sensitivity compared to voltmeters.

Question 21:
Explain the principle of a potentiometer and derive the expression for the comparison of emf of two primary cells using it. Discuss the advantages of a potentiometer over a voltmeter.
Answer:

The potentiometer is a device used to measure the emf (electromotive force) of a cell or compare the emfs of two cells accurately without drawing any current from them. It works on the principle that the potential difference across a uniform wire is directly proportional to its length when a constant current flows through it.


Principle: When a steady current flows through a uniform wire of length L, the potential drop V across a segment of length l is given by V = k * l, where k is the potential gradient.


Derivation for comparing emfs of two cells:
1. Connect the two primary cells (E1 and E2) in series opposition and balance them against the potentiometer wire.
2. For the first cell (E1), the balancing length is l1, so E1 = k * l1.
3. For the second cell (E2), the balancing length is l2, so E2 = k * l2.
4. Dividing the two equations, we get: E1/E2 = l1/l2.


Advantages of potentiometer over voltmeter:

  • Potentiometer measures the emf directly without drawing any current from the cell, while a voltmeter draws some current, leading to potential drop due to internal resistance.
  • It provides higher accuracy as it uses null deflection method.
  • It can measure small potential differences precisely.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A student measures the current-voltage characteristics of a silicon diode and plots the graph. The graph shows a sharp rise in current beyond 0.7 V. Explain the depletion region and breakdown voltage in this context.
Answer:
Case Deconstruction

The diode's behavior is due to the depletion region, where electron-hole pairs recombine, creating a barrier. Beyond 0.7 V, the barrier is overcome, allowing current flow.

Theoretical Application
  • Silicon diodes have a threshold voltage of ~0.7 V due to their bandgap.
  • Breakdown voltage occurs when reverse bias exceeds a limit, causing avalanche multiplication.
Critical Evaluation

Our textbook shows that excessive voltage can damage the diode. For example, Zener diodes exploit breakdown for voltage regulation.

Question 2:
Two resistors of 2 Ω and 4 Ω are connected in parallel to a 12 V battery. Calculate the equivalent resistance and power dissipated in the 4 Ω resistor.
Answer:
Case Deconstruction

Parallel resistors follow 1/Req = 1/R1 + 1/R2. Power is P = V2/R.

Theoretical Application
  • Req = (2×4)/(2+4) = 1.33 Ω.
  • Power in 4 Ω: P = 122/4 = 36 W.
Critical Evaluation

We studied that parallel circuits divide current, but voltage remains equal. For example, household wiring uses parallel connections for uniform voltage.

Question 3:
A potentiometer wire of length 10 m has a resistance of 20 Ω. It is connected to a 5 V battery. Discuss how the null point shifts if the wire is replaced with a thicker one of the same material.
Answer:
Case Deconstruction

Thicker wire reduces resistance per unit length, altering the null point due to changed potential gradient.

Theoretical Application
  • Resistance R ∝ 1/A (A = cross-section area).
  • Null point shifts toward the longer end to balance EMF.
Critical Evaluation

Our textbook shows that potentiometer accuracy depends on wire uniformity. For example, copper wires are preferred for low resistivity.

Question 4:
A 10 μF capacitor is charged to 100 V and discharged through a 1 kΩ resistor. Analyze the time constant and energy dissipated during discharge.
Answer:
Case Deconstruction

The time constant (τ = RC) determines discharge rate. Energy is stored as ½CV2.

Theoretical Application
  • τ = 1×103 × 10×10-6 = 0.01 s.
  • Energy dissipated: ½ × 10-5 × 1002 = 0.05 J.
Critical Evaluation

We studied that 63% charge decays in one τ. For example, camera flashes use RC circuits for timed discharges.

Question 5:
A student sets up a circuit with a nichrome wire of length 2m and cross-sectional area 0.5 mm² connected to a 6V battery. The current measured is 1.5A. Analyze the resistivity of nichrome and compare it with copper (ρ = 1.68 × 10⁻⁸ Ωm).
Answer:
Case Deconstruction

Given: V = 6V, I = 1.5A, L = 2m, A = 0.5 × 10⁻⁶ m². Using Ohm's Law (R = V/I), R = 4Ω.

Theoretical Application

Resistivity (ρ) = RA/L = (4Ω × 0.5 × 10⁻⁶ m²)/2m = 1 × 10⁻⁶ Ωm. Copper's resistivity (1.68 × 10⁻⁸ Ωm) is significantly lower, confirming nichrome's higher resistance.

Critical Evaluation

Nichrome's high ρ makes it suitable for heating elements, unlike copper. Our textbook shows similar calculations for tungsten (ρ = 5.6 × 10⁻⁸ Ωm), reinforcing material-based resistivity trends.

Question 6:
In a lab, two identical semiconductor diodes are connected in series with a 5V battery. One diode shows a voltage drop of 0.7V, while the other shows 0.3V. Critically evaluate this observation using the concept of dynamic resistance.
Answer:
Case Deconstruction

Diodes in series should ideally have equal voltage drops. The discrepancy (0.7V vs. 0.3V) suggests non-identical dynamic resistance due to manufacturing variations.

Theoretical Application

Dynamic resistance (r_d = ΔV/ΔI) varies with current. Our textbook shows that r_d decreases exponentially with forward bias, explaining the lower drop (0.3V) for the diode with higher current.

Critical Evaluation

This highlights real-world deviations from ideal diode models. Examples like LED forward voltages (1.8V-3.3V) further validate material-dependent behavior.

Question 7:
A potentiometer wire of length 10m and resistance 20Ω is used to measure the EMF of a cell. The balancing length is 4m. Calculate the cell's EMF if the driver cell has 2V. Justify why potentiometers are preferred over voltmeters for EMF measurement.
Answer:
Case Deconstruction

Given: L = 10m, R = 20Ω, balancing length (l) = 4m, driver EMF = 2V. Potential gradient (k) = 2V/10m = 0.2V/m. Cell EMF = k × l = 0.8V.

Theoretical Application

Potentiometers measure EMF without drawing current, eliminating internal resistance errors. Voltmeters alter circuit conditions, as shown in our textbook's comparison of open-circuit vs. loaded measurements.

Critical Evaluation

This method's precision is evident in standard cell calibrations. Example: Weston cell EMF (1.0186V) is measured exclusively using potentiometers.

Question 8:
A carbon resistor with color codes Brown-Black-Red-Gold shows a measured resistance of 980Ω. Analyze the tolerance deviation and discuss the implications for high-precision circuits.
Answer:
Case Deconstruction

Color code: Brown (1), Black (0), Red (10²), Gold (±5%). Nominal value = 10 × 10² = 1kΩ. Measured 980Ω is within tolerance (950Ω-1050Ω).

Theoretical Application

Tolerance reflects manufacturing precision. Our textbook shows military-grade resistors (±1%) outperform consumer-grade (±5%) in critical applications like medical equipment.

Critical Evaluation

For circuits needing ±0.1% precision (e.g., quantum experiments), carbon resistors are inadequate. Examples like strain gauges require metal-film resistors for stability.

Question 9:
A student measures the current-voltage (I-V) characteristics of a semiconductor diode and plots the graph. The graph shows non-linear behavior. Explain why the diode exhibits this behavior and compare it with an ohmic conductor.
Answer:
Case Deconstruction

The diode's non-linear I-V curve arises due to its p-n junction, which allows current only in forward bias after overcoming the barrier potential.

Theoretical Application
  • Ohmic conductors (e.g., copper) follow Ohm's Law (V ∝ I) due to constant resistance.
  • Diodes have dynamic resistance, varying with applied voltage.
Critical Evaluation

Our textbook shows silicon diodes have ~0.7V threshold. This non-linearity enables applications like rectifiers, unlike ohmic materials used in wires.

Question 10:
In a circuit, two resistors of 3Ω and 6Ω are connected in parallel. A 2A current enters the combination. Calculate the current through each resistor and justify your steps using Kirchhoff's laws.
Answer:
Case Deconstruction

Parallel resistors share voltage but divide current inversely to resistance (I ∝ 1/R).

Theoretical Application
  • Equivalent resistance: 1/Req = 1/3 + 1/6 → Req = 2Ω.
  • Voltage across combination: V = IR = 2A × 2Ω = 4V.
  • Currents: I3Ω = 4V/3Ω ≈ 1.33A, I6Ω = 4V/6Ω ≈ 0.67A.
Critical Evaluation

Kirchhoff's current law is satisfied (1.33A + 0.67A = 2A), validating our calculations.

Question 11:
A potentiometer wire of length 10m has a resistance of 20Ω. It is connected to a standard cell of emf 2V. Determine the potential gradient and explain how it helps in measuring unknown emf accurately.
Answer:
Case Deconstruction

Potential gradient (k) is voltage drop per unit length, crucial for null-point measurements.

Theoretical Application
  • Current in wire: I = V/R = 2V/20Ω = 0.1A.
  • Potential gradient: k = IR/L = (0.1A × 20Ω)/10m = 0.2V/m.
Critical Evaluation

Our textbook shows potentiometers eliminate internal resistance errors, unlike voltmeters. For an unknown emf E, balance length l gives E = kl.

Question 12:
The resistivity of a wire decreases by 2% when its temperature increases by 10°C. Analyze whether the material is a conductor or semiconductor and support your answer with band theory.
Answer:
Case Deconstruction

Resistivity-temperature relation distinguishes conductors (ρ ↑ with T) from semiconductors (ρ ↓ with T).

Theoretical Application
  • Conductors: Increased lattice vibrations scatter electrons, raising ρ.
  • Semiconductors: More electrons jump to conduction band, reducing ρ.
Critical Evaluation

Since ρ decreases, it matches semiconductors like silicon. Band theory explains this as thermal energy overcomes the band gap, increasing charge carriers.

Question 13:

A student sets up an experiment to verify Ohm's Law using a resistor, battery, ammeter, and voltmeter. The observed readings are:

  • Voltmeter (V): 2V, 4V, 6V
  • Ammeter (I): 0.5A, 1.0A, 1.5A

However, when plotting the V-I graph, the student notices the line does not pass through the origin. Explain the possible reason for this deviation and how it affects the resistance calculation.

Answer:

The deviation in the V-I graph not passing through the origin can occur due to:

  • Zero error in the ammeter or voltmeter, leading to incorrect initial readings.
  • Presence of internal resistance in the battery, causing a voltage drop even when current is zero.

This affects resistance calculation as:


1. The slope of the V-I graph still gives the resistance, but the intercept indicates an additional voltage loss.


2. If the internal resistance (r) is significant, the actual resistance (R) of the resistor must be calculated using V = E - Ir, where E is the emf of the battery.


To minimize errors, the student should calibrate instruments and account for internal resistance.

Question 14:

In a household circuit, a fuse of rating 5A is connected to a device drawing 4A current. The device has a resistance of 55Ω. Suddenly, the current surges to 6A due to a fault. Explain the role of the fuse in this scenario and calculate the power dissipated before and after the fault.

Answer:

The fuse acts as a safety device by melting and breaking the circuit when current exceeds its rating (5A).


Before fault:
Current (I) = 4A
Resistance (R) = 55Ω
Power (P) = I²R = (4)² × 55 = 16 × 55 = 880W


After fault:
Current surges to 6A (>5A), so the fuse melts, cutting off the circuit.
If the fuse were absent:
Power (P) = (6)² × 55 = 36 × 55 = 1980W, which could overheat wires and cause fire hazards.


Thus, the fuse prevents excessive power dissipation and potential damage.

Question 15:

A student sets up an experiment to verify Ohm's Law using a resistor, a battery, an ammeter, and a voltmeter. The voltmeter reading shows 2.5 V, and the ammeter shows 0.5 A. Later, the student replaces the resistor with another one of unknown resistance and observes the voltmeter reading as 3.0 V and the ammeter as 0.3 A.

(a) Calculate the resistance of the first resistor.
(b) Determine the resistance of the unknown resistor.
(c) State whether the material of the unknown resistor obeys Ohm's Law. Justify your answer.

Answer:

(a) Using Ohm's Law (V = IR), the resistance of the first resistor is calculated as:
R₁ = V/I = 2.5 V / 0.5 A = 5 Ω.

(b) For the unknown resistor:
R₂ = V/I = 3.0 V / 0.3 A = 10 Ω.

(c) The unknown resistor obeys Ohm's Law because the ratio V/I remains constant (10 Ω) for different readings, indicating a linear relationship between voltage and current.

Note: Ohm's Law is valid only for ohmic conductors where resistance remains constant with changing voltage.

Question 16:

In a household circuit, two appliances—a 100 W bulb and a 500 W electric iron—are connected in parallel to a 220 V supply. The circuit has a fuse rated at 5 A.

(a) Calculate the current drawn by each appliance.
(b) Determine the total current in the circuit.
(c) Will the fuse blow? Justify your answer.

Answer:

(a) Using the formula P = VI, the current drawn by each appliance is:
For the bulb: I₁ = P/V = 100 W / 220 V ≈ 0.45 A.
For the iron: I₂ = P/V = 500 W / 220 V ≈ 2.27 A.

(b) Total current in the parallel circuit is the sum of individual currents:
Itotal = I₁ + I₂ ≈ 0.45 A + 2.27 A = 2.72 A.

(c) The fuse will not blow because the total current (2.72 A) is less than the fuse rating (5 A). However, if more appliances are added, exceeding 5 A, the fuse will blow to protect the circuit.

Note: Fuses are safety devices that prevent overheating and fire hazards in circuits.

Question 17:
A student sets up an experiment to verify Ohm's Law using a resistor, a battery, an ammeter, and a voltmeter. The student records the following observations:

Voltage (V): 1.0, 2.0, 3.0, 4.0
Current (I): 0.2, 0.4, 0.6, 0.8

(a) Plot a graph of V vs I and determine the resistance of the resistor.
(b) If the student accidentally connects the ammeter in parallel with the resistor, what would be the observed effect on the readings? Justify your answer.
Answer:

(a) To plot the graph of V vs I, follow these steps:
1. Take Voltage (V) on the y-axis and Current (I) on the x-axis.
2. Plot the given data points: (0.2, 1.0), (0.4, 2.0), (0.6, 3.0), (0.8, 4.0).
3. Draw a straight line passing through the origin and the plotted points.
4. The slope of the line (V/I) gives the resistance (R).

Calculation:
Slope = ΔV/ΔI = (4.0 - 1.0) / (0.8 - 0.2) = 3.0 / 0.6 = 5.0 Ω.
Thus, the resistance of the resistor is 5.0 Ω.

(b) If the ammeter is connected in parallel with the resistor:
1. The ammeter has very low resistance, so most of the current will flow through it instead of the resistor.
2. The voltmeter will still measure the correct voltage across the resistor, but the ammeter will show an abnormally high current.
3. This will lead to incorrect calculations, as Ohm's Law (V = IR) won't hold true for the resistor alone.
4. The circuit may also get damaged due to excessive current flow through the ammeter.

Question 18:
In a household circuit, a fuse wire of rating 5A is used to protect an appliance. The appliance has a power rating of 1.1 kW and operates at 220V.

(a) Calculate the current drawn by the appliance under normal operation. Is the fuse wire suitable?
(b) Explain what happens if a thicker fuse wire of rating 15A is used instead. What safety risk does this pose?
Answer:

(a) To calculate the current drawn by the appliance:
Given: Power (P) = 1.1 kW = 1100 W, Voltage (V) = 220 V.
Using the formula P = VI,
I = P / V = 1100 / 220 = 5 A.

The fuse wire is rated at 5A, which matches the current drawn by the appliance. Hence, it is suitable as it will blow if the current exceeds 5A, protecting the appliance.

(b) If a thicker fuse wire of 15A is used:
1. The fuse will not blow even if the current exceeds 5A (up to 15A).
2. This defeats the purpose of the fuse, as the appliance may overheat or get damaged due to excessive current.
3. It poses a fire hazard because the wiring may overheat, leading to short circuits or insulation damage.
4. Always use a fuse with a rating slightly higher than the normal operating current, but not excessively high, to ensure safety.

Question 19:
A student sets up an experiment to determine the internal resistance of a cell using a potentiometer. The balancing length is found to be 560 cm when the cell is in an open circuit. When a standard resistor of 5 Ω is connected across the cell, the balancing length reduces to 400 cm.

Based on this case, answer the following:

  • Calculate the internal resistance of the cell.
  • Explain why the balancing length decreases when the resistor is connected.
Answer:

To calculate the internal resistance (r) of the cell, we use the formula:
r = R (L₁ - L₂) / L₂
where:
R = 5 Ω (standard resistor)
L₁ = 560 cm (open circuit balancing length)
L₂ = 400 cm (balancing length with resistor connected)

Substituting the values:
r = 5 × (560 - 400) / 400
r = 5 × 160 / 400
r = 2 Ω

The balancing length decreases because connecting the resistor causes a potential drop across the internal resistance of the cell. This reduces the terminal voltage available to the potentiometer, leading to a shorter balancing length.

Question 20:
In a laboratory, a student measures the current-voltage (I-V) characteristics of a nichrome wire at different temperatures. The graph shows a linear relationship at lower voltages but deviates at higher voltages.

Based on this case, answer the following:

  • Why does the I-V graph become non-linear at higher voltages?
  • How does temperature affect the resistance of the nichrome wire?
Answer:

The I-V graph becomes non-linear at higher voltages because:

  • The nichrome wire heats up significantly due to Joule heating, increasing its temperature.
  • As temperature rises, the resistance of the wire increases because the resistivity of nichrome is temperature-dependent.

Temperature affects the resistance as follows:
Nichrome has a positive temperature coefficient, meaning its resistance increases with temperature.
Mathematically, R = R₀ (1 + αΔT), where α is the temperature coefficient.
Higher temperatures cause more lattice vibrations, impeding electron flow and increasing resistance.

Question 21:
A student sets up an experiment to determine the internal resistance of a cell using a potentiometer. The balancing length is found to be 560 cm when the cell is in an open circuit. When a standard resistor of 2 Ω is connected across the cell, the balancing length reduces to 400 cm.

Based on this case, answer the following:

  • Calculate the internal resistance of the cell.
  • Explain why the balancing length decreases when the resistor is connected.
Answer:

To calculate the internal resistance (r) of the cell, we use the formula:
r = R (L₁ - L₂) / L₂
where:
R = Standard resistor (2 Ω)
L₁ = Balancing length in open circuit (560 cm)
L₂ = Balancing length with resistor (400 cm)
Substituting the values:
r = 2 Ω × (560 cm - 400 cm) / 400 cm
r = 2 Ω × (160 cm) / 400 cm
r = 0.8 Ω.

The balancing length decreases because connecting the resistor causes a potential drop across the internal resistance of the cell. This reduces the terminal voltage available to the potentiometer, leading to a shorter balancing length. The decrease in voltage is due to energy dissipation within the cell itself.

Question 22:
In a laboratory, a group of students measures the resistivity of a given wire using a meter bridge. They obtain the following data:
  • Balancing length (L) = 45 cm
  • Known resistance (R) = 10 Ω
  • Diameter of the wire (d) = 0.4 mm

Answer the following:

  • Calculate the unknown resistance (S) of the wire.
  • Determine the resistivity (ρ) of the wire material.
Answer:

Using the meter bridge formula for unknown resistance (S):
S = R (100 - L) / L
Substituting the values:
S = 10 Ω × (100 cm - 45 cm) / 45 cm
S = 10 Ω × 55 cm / 45 cm
S ≈ 12.22 Ω.

To find the resistivity (ρ), we use:
ρ = (S × πd²) / (4l)
where l is the length of the wire (assumed as 1 m for standard lab setup).
Given diameter (d) = 0.4 mm = 0.0004 m,
ρ = (12.22 Ω × 3.14 × (0.0004 m)²) / (4 × 1 m)
ρ ≈ 1.53 × 10⁻⁶ Ω·m.

This value indicates the wire's material has moderate resistivity, suitable for laboratory experiments.

Chat on WhatsApp