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Samacheer Kalvi 12th Physics Chapter 2 Exercise Solutions

Tamil Nadu Class 12 Physics Chapter 2 Current Electricity exercise solutions with Ohm law, resistance, Kirchhoff laws, Wheatstone bridge, potentiometer, and numerical steps.

Updated 25 May 202617 min read

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

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Chapter 2 — Current Electricity Exercise Solutions

Circuit problems become easier when you redraw the circuit and mark current direction before writing equations.

How to Use This Page

  • Start with the Key Concepts table to quickly revise definitions before attempting problems.
  • Use the Formula Recap section as a ready reference while solving numericals.
  • Work through MCQs and short answers first to test recall, then move to worked examples and detailed answers.
  • After each practice session, check the Common Mistakes section to catch errors you may have made.
  • Before the exam, go through the Exam Checklist to ensure no topic is left uncovered.

Key Concepts (Quick Recap)

TermDefinition
Electric currentRate of flow of electric charge through a conductor; I = Q/t, unit ampere (A)
Drift velocityAverage velocity with which free electrons drift in a conductor under an electric field; vd = eEτ/m
Ohm's lawCurrent through a conductor is directly proportional to the potential difference across it, provided temperature remains constant; V = IR
ResistanceProperty of a conductor that opposes the flow of current; depends on material, length, and cross-sectional area; R = ρl/A
ResistivityIntrinsic property of a material that determines its resistance per unit length and area; depends on material and temperature, not dimensions
Series combinationResistors connected end-to-end; same current flows through each; equivalent resistance is the sum of individual resistances
Parallel combinationResistors connected between the same two points; same potential difference across each; reciprocal of equivalent resistance equals sum of reciprocals
Kirchhoff's lawsJunction rule (algebraic sum of currents at a junction is zero) and Loop rule (algebraic sum of potential changes around a closed loop is zero)
Wheatstone bridgeArrangement of four resistors forming a diamond; balanced when P/Q = R/S, no current through galvanometer
PotentiometerDevice to measure potential difference using null deflection; more accurate than voltmeter because it draws no current at balance
Internal resistanceResistance offered by the electrolyte inside a cell; causes terminal voltage to be less than EMF; V = EMF − Ir
Electrical powerRate at which electrical energy is consumed or dissipated; P = VI = I²R = V²/R, unit watt (W)
EMFElectromotive force; the maximum potential difference a cell can provide when no current is drawn
SuperconductivityPhenomenon where resistance of certain materials drops to zero below a critical temperature

Formula Recap

ConceptFormula
Ohm lawV = IR
ResistanceR = rho l / A
Series resistanceR = R1 + R2 + ...
Parallel resistance1/R = 1/R1 + 1/R2 + ...
Electric powerP = VI = I²R = V²/R
Drift currentI = neAvd

Kirchhoff Law Pattern

1. Assume current directions. 2. Apply junction rule: incoming current = outgoing current. 3. Apply loop rule: algebraic sum of potential changes is zero. 4. Solve the simultaneous equations. 5. If current is negative, actual direction is opposite.

Short Answer Points

  • Ohm's law is valid only when physical conditions remain constant.
  • Resistivity depends on material and temperature.
  • Potentiometer is more accurate than voltmeter because it uses null deflection.
  • Wheatstone bridge is balanced when P/Q = R/S.

I. Choose the Best Answer (MCQs)

1. A conductor carries a current of 2 A. The charge flowing through it in 5 seconds is:

(A) 5 C    (B) 10 C    (C) 2.5 C    (D) 0.4 C

Answer(B) 10 C — Q = It = 2 × 5 = 10 C.

2. According to Ohm's law, the V-I graph for a metallic conductor is:

(A) A parabola    (B) A straight line passing through origin    (C) A circle    (D) A hyperbola

Answer(B) A straight line passing through origin — Ohm's law states V ∝ I, giving a linear graph through the origin.

3. Two resistors of 6 Ω and 3 Ω are connected in parallel. The equivalent resistance is:

(A) 9 Ω    (B) 4.5 Ω    (C) 2 Ω    (D) 1 Ω

Answer(C) 2 Ω — 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so R = 2 Ω.

4. In a Wheatstone bridge, the galvanometer shows null deflection when:

(A) P × S = Q × R    (B) P/Q = R/S    (C) P × R = Q × S    (D) Both (A) and (B)

Answer(D) Both (A) and (B) — P × S = Q × R is the cross-product form; rearranged, P/Q = R/S is the ratio form.

5. A potentiometer is preferred over a voltmeter for measuring EMF because:

(A) It is cheaper    (B) It draws no current at balance    (C) It is portable    (D) It has a digital display

Answer(B) It draws no current at balance — At null point, the potentiometer draws zero current, so no voltage drop across internal resistance.

6. The internal resistance of an ideal cell is:

(A) Zero    (B) Infinite    (C) 1 Ω    (D) Depends on temperature

Answer(A) Zero — An ideal cell has no internal resistance; all EMF appears as terminal voltage.

7. The SI unit of resistivity is:

(A) Ω    (B) Ω m    (C) Ω/m    (D) m/Ω

Answer(B) Ω m — From R = ρl/A, ρ = RA/l, so unit is Ω·m²/m = Ω m.

8. If the length of a wire is doubled, its resistance:

(A) Remains same    (B) Doubles    (C) Becomes half    (D) Becomes four times

Answer(B) Doubles — R = ρl/A; if l doubles with same A and ρ, R doubles.

9. The drift velocity of electrons in a conductor is proportional to:

(A) Electric field    (B) Square of electric field    (C) Cube of electric field    (D) Independent of electric field

Answer(A) Electric field — vd = eEτ/m, so vd ∝ E.

10. Two identical cells each of EMF 1.5 V and internal resistance 0.5 Ω are connected in parallel. The total EMF and internal resistance are:

(A) 3 V, 1 Ω    (B) 1.5 V, 0.25 Ω    (C) 1.5 V, 1 Ω    (D) 3 V, 0.25 Ω

Answer(B) 1.5 V, 0.25 Ω — Parallel cells of same EMF: total EMF = 1.5 V, total internal resistance = 0.5/2 = 0.25 Ω.

II. Fill in the Blanks

1. The SI unit of electric current is __________. Ampere

2. According to Ohm's law, V = __________. IR

3. The resistance of a conductor is directly proportional to its __________. Length

4. In a Wheatstone bridge, the balance condition is P/Q = __________. R/S

5. The potentiometer works on the principle of __________ deflection. Null

6. The terminal voltage of a cell is less than its EMF due to __________ resistance. Internal

7. When resistors are connected in series, the __________ is the same through each resistor. Current

8. The reciprocal of resistance is called __________. Conductance

III. True or False (with Corrections)

1. Statement: Ohm's law is valid for all types of conductors. False. Ohm's law is valid only for ohmic conductors (metallic conductors at constant temperature). It does not hold for semiconductors, electrolytes, or diodes.

2. Statement: In a parallel combination, the same current flows through each resistor. False. In parallel, the same potential difference appears across each resistor, but the current through each resistor may differ depending on its resistance.

3. Statement: The potentiometer draws current from the circuit at the balance point. False. At the balance (null) point, the potentiometer draws zero current, which is why it gives more accurate EMF measurements than a voltmeter.

4. Statement: Drift velocity of electrons in a conductor is very high, close to the speed of light. False. Drift velocity is very small, typically of the order of 10⁻⁵ m/s. The electric signal propagates at near light speed, but individual electrons drift slowly.

5. Statement: A Wheatstone bridge can be used to measure an unknown resistance accurately. True. The Wheatstone bridge is one of the most accurate methods for measuring unknown resistances using the null deflection principle.

6. Statement: Internal resistance of a cell increases with the age of the cell. True. As a cell ages, the electrolyte weakens and electrodes degrade, increasing the internal resistance and reducing the terminal voltage.

7. Statement: Superconductors have very high resistance at room temperature. False. Superconductors have zero resistance below their critical temperature. At room temperature they may have resistance, but the defining property is zero resistance below Tc.

IV. Match the Following

Column AColumn B
1. OhmA. Rate of flow of charge
2. AmpereB. V = IR
3. ResistivityC. Property opposing current flow
4. ResistanceD. Depends on material, not dimensions
5. PotentiometerE. Null deflection method
6. Internal resistanceF. Causes terminal voltage < EMF
Answer1–B, 2–A, 3–D, 4–C, 5–E, 6–F

V. Answer Briefly (2-mark)

1. State Ohm's law and mention the condition under which it is valid.
Answer

Ohm's law states that the current through a metallic conductor is directly proportional to the potential difference across it, provided the temperature and other physical conditions remain constant. Mathematically, V = IR, where R is the resistance. It is valid only for ohmic conductors at constant temperature and does not apply to semiconductors, diodes, or electrolytes.

2. Why is a potentiometer preferred over a voltmeter for measuring EMF?
Answer

A potentiometer is preferred because it uses the null deflection method — at the balance point, it draws no current from the cell being measured. A voltmeter draws some current, causing a voltage drop across the internal resistance and giving a reading less than the actual EMF. The potentiometer thus provides a true EMF measurement without loading the circuit.

3. What happens to the resistance of a metallic conductor when temperature increases?
Answer

The resistance of a metallic conductor increases with temperature. As temperature rises, the atoms in the conductor vibrate more vigorously, increasing the frequency of collisions with free electrons. This reduces the drift velocity and increases resistance. The relation is Rt = R0(1 + αΔT), where α is the temperature coefficient of resistance.

4. Distinguish between EMF and terminal voltage of a cell.
Answer

EMF is the maximum potential difference a cell can provide when no current is drawn (open circuit). Terminal voltage is the actual potential difference across the cell's terminals when current flows. Terminal voltage is less than EMF due to the voltage drop across internal resistance: V = EMF − Ir. Only when no current flows (I = 0) does terminal voltage equal EMF.

5. What is the principle of a Wheatstone bridge?
Answer

The Wheatstone bridge works on the principle of null deflection. Four resistors P, Q, R, and S are arranged in a diamond with a galvanometer between the midpoints. When the ratio P/Q equals R/S, no current flows through the galvanometer and the bridge is said to be balanced. This null condition allows precise measurement of an unknown resistance.

Worked Examples

Example 1 — Resistors in parallel

Find the equivalent resistance of 2 Ω, 3 Ω, and 6 Ω in parallel.

1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 R = 1 Ω

Example 2 — Ohm's law with internal resistance

A cell of EMF 12 V and internal resistance 0.5 Ω drives a 5.5 Ω resistor. Find the current.

I = EMF / (R + r) = 12 / (5.5 + 0.5) = 12 / 6 = 2 A

Example 3 — Power dissipated

Find the power in a 10 Ω resistor carrying 3 A.

P = I²R = (3)² × 10 = 9 × 10 = 90 W

VI. Answer in Detail (5-mark)

1. State and explain Kirchhoff's laws with an illustration.
Answer

Kirchhoff's laws are two fundamental rules for analysing complex electrical circuits. The first law, called the junction rule, states that the algebraic sum of currents meeting at a junction is zero. This is based on the conservation of charge — whatever current enters a junction must leave it. If three wires meet at a point with currents I1, I2 entering and I3 leaving, then I1 + I2 − I3 = 0.

The second law, called the loop rule, states that the algebraic sum of potential changes around any closed loop in a circuit is zero. This is based on the conservation of energy. While traversing a loop, a rise in potential across a battery (from negative to positive) is taken as positive, and a drop across a resistor (in the direction of current) is taken as negative. For a simple loop with a battery of EMF E and two resistors R1 and R2 carrying current I, the loop equation is E − IR1 − IR2 = 0.

To apply these laws, first assign current directions to each branch, write junction equations at each node, then write loop equations for each independent loop. Solve the simultaneous equations for unknown currents. If a current comes out negative, the assumed direction was opposite.

2. Derive the condition for a balanced Wheatstone bridge and explain its applications.
Answer

A Wheatstone bridge consists of four resistors P, Q, R, and S connected in a diamond pattern. A battery is connected between two opposite points, and a galvanometer between the other two. Let current I enter at point A and split into I1 through P and I2 through R. At the junction between P and R, current I1 goes through Q and I2 goes through S.

For the bridge to be balanced, no current flows through the galvanometer. This means the potential at the two ends of the galvanometer must be equal. Let this common potential be applied. Using the junction rule, the current through Q is also I1 and through S is also I2.

Writing the potential difference across P and R: I1 × P = I2 × R. Writing the potential difference across Q and S: I1 × Q = I2 × S. Dividing these equations: P/Q = R/S, or cross-multiplying, P × S = Q × R. This is the balance condition.

Applications include measuring unknown resistance accurately, measuring small changes in resistance using a metre bridge, and strain gauge measurements. The null deflection method eliminates errors from instrument calibration.

3. Explain the principle and working of a potentiometer. How is it used to compare EMFs of two cells?
Answer

A potentiometer is a device that measures potential difference using the null deflection method. It consists of a long uniform resistance wire AB of length L, through which a steady current flows from a driver cell. A sliding contact (jockey) can tap the wire at any point, and the potential difference across a length l of the wire is proportional to that length: V = kl, where k is the potential gradient (voltage per unit length).

To compare the EMFs of two cells E1 and E2, first connect cell E1 in the secondary circuit with its positive terminal connected to end A. Slide the jockey to find the null point at length l1. Then E1 = k × l1. Next, replace E1 with cell E2 and find the new null point at length l2. Then E2 = k × l2. The ratio E1/E2 = l1/l2. Since both measurements use the same potential gradient, the value of k cancels out.

The potentiometer is superior to a voltmeter because at the null point it draws no current from the cell, so there is no voltage drop across internal resistance. This gives the true EMF. It is used in laboratories for calibrating ammeters and voltmeters, comparing resistances, and measuring internal resistance of cells.

4. Derive the relation between drift velocity and electric current in a conductor.
Answer

Consider a conductor of cross-sectional area A carrying current I. Let n be the number of free electrons per unit volume and e be the charge on each electron. When an electric field E is applied along the conductor, free electrons drift with an average velocity vd called the drift velocity.

In time Δt, each electron travels a distance vd × Δt. Consider a volume element of the conductor of length vd × Δt. The volume of this element is A × vd × Δt. The number of free electrons in this volume is n × A × vd × Δt. The total charge crossing a cross-section in time Δt is ΔQ = n × A × vd × Δt × e.

Current is the rate of flow of charge: I = ΔQ/Δt = neAvd. This is the fundamental relation connecting current and drift velocity. Rearranging, vd = I/(neA), which shows that drift velocity is inversely proportional to the cross-sectional area for a given current.

The drift velocity is related to the electric field by vd = eEτ/m, where τ is the average relaxation time and m is the mass of an electron. Combining both equations gives I = neA(eEτ/m) = (ne²τA/m)E, from which we can derive the resistivity ρ = m/(ne²τ).

Common Mistakes to Avoid

1. ❌ Adding parallel resistances directly: R1 + R2 ✅ Use reciprocal formula: 1/R = 1/R1 + 1/R2, then invert the result.

2. ❌ Forgetting internal resistance when calculating current from a cell ✅ Always use I = EMF / (R + r). The total resistance includes both external and internal resistance.

3. ❌ Wrong sign convention in Kirchhoff's loop equation ✅ Choose a traversal direction and stick to it. EMF is positive when traversed from negative to positive; potential drop across a resistor is negative when traversed in the direction of current.

4. ❌ Not converting milliampere to ampere or milli-volt to volt before substituting ✅ Always express quantities in SI units: ampere, volt, ohm, second before solving.

5. ❌ Assuming current is the same in all branches of a parallel circuit ✅ In parallel, voltage is the same across each branch but current divides inversely proportional to resistance.

6. ❌ Confusing resistivity with resistance ✅ Resistivity (ρ) is a material property depending on material and temperature. Resistance (R = ρl/A) also depends on geometry (length and area). Two wires of the same material can have different resistances but the same resistivity.

Exam Checklist

  • [ ] Can you state Ohm's law and the condition for its validity?
  • [ ] Can you derive the relation between current and drift velocity (I = neAvd)?
  • [ ] Can you calculate equivalent resistance for series and parallel combinations?
  • [ ] Can you apply Kirchhoff's junction rule and loop rule to write equations for a circuit?
  • [ ] Can you derive the Wheatstone bridge balance condition (P/Q = R/S)?
  • [ ] Can you explain why a potentiometer is more accurate than a voltmeter?
  • [ ] Can you solve problems involving internal resistance and terminal voltage (V = EMF − Ir)?
  • [ ] Do you convert all quantities to SI units before substituting into formulas?
  • [ ] Can you calculate electrical power using P = VI, P = I²R, and P = V²/R?
  • [ ] Can you distinguish between EMF, terminal voltage, and potential difference?

Frequently Asked Questions

Equivalent resistance, Kirchhoff law circuit problems, meter bridge, potentiometer, and internal resistance numericals are important.