CBSE Class 12 Physics Electric Circuits Important Questions

CBSE Class 12 Physics Electric Circuits Important Questions. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

CLASS 12 PHYSICS : ELECTRIC CIRCUITS
MISCELLANEOUS ASSIGNMENT

Comprehension-1

Two persons are pulling a square of side a along one of the diameters horizontally to make it rhombus. Plane of rhombus is always vertical and uniform magnetic field B exist perpendicular to plane. They start pulling at t = 0 and with constant velocity v.

1. The induced emf in the frame when angle at corner being pulled is 60°
(a) Bav
(b) 2Bav
(c) \( \frac{Bav}{2} \)
(d) \( \frac{Bav}{4} \)

Answer: (b) 2Bav

1. Let θ be the angle at a pulled corner. The diagonal along the pull is d = 2a cos(θ/2), and each person pulls with speed v, so dd/dt = 2v.
2. The area of the rhombus is A = a² sin θ, so the emf = B \( \frac{dA}{dt} \) = B a² cos θ \( \frac{dθ}{dt} \).
3. From d = 2a cos(θ/2): -a sin(θ/2) \( \frac{dθ}{dt} \) = 2v, so the magnitude of \( \frac{dθ}{dt} \) = \( \frac{2v}{a sin(θ/2)} \).
4. emf = \( \frac{2Bav cos θ}{sin(θ/2)} \); at θ = 60°, emf = \( \frac{2Bav \times 0.5}{0.5} \) = 2Bav.
5. Hence, the correct option is (b).

Teacher's Note:
a) The emf is the rate of change of magnetic flux, so write the area of the rhombus in terms of the corner angle.
b) Both pullers move with speed v, so the diagonal lengthens at 2v, not v.

2. If the resistance of the frame is R the current induced is
(a) \( \frac{Bav}{2R} \)
(b) \( \frac{2Bav}{R} \)
(c) \( \frac{Bav}{R} \)
(d) \( \frac{Bav}{4R} \)

Answer: (b) \( \frac{2Bav}{R} \)

1. Induced current = \( \frac{emf}{R} \).
2. From the previous question, emf = 2Bav at θ = 60°.
3. Current = \( \frac{2Bav}{R} \).
4. Hence, the correct option is (b).

Teacher's Note:
a) Use Ohm's law I = emf/R with the emf found in the previous question.
b) This holds only at the instant when the pulled corner angle is 60°.

3. Finally square frame reduces to straight wire. The total charge flown is
(a) \( \frac{a^{2} B}{R} \)
(b) \( \frac{a^{2} B}{2R} \)
(c) \( \frac{2a^{2} B}{R} \)
(d) \( \frac{4a^{2} B}{R} \)

Answer: (a) \( \frac{a^{2} B}{R} \)

1. Charge flown q = \( \frac{Δφ}{R} \) = \( \frac{B ΔA}{R} \).
2. Initial area = a² and final area = 0 (the frame becomes a straight wire), so ΔA = a².
3. q = \( \frac{a^{2} B}{R} \).
4. Hence, the correct option is (a).

Teacher's Note:
a) Total charge depends only on the change of flux and the resistance, not on the time taken.
b) The flux is zero at the end because the area has shrunk to zero.

Comprehension-2

In case of analysis of circuits, containing cells, resistences and inductances two things are very important one is conservation of charge which leads to the fact that at any junction of circuit incoming current is equal to out going current. The other thing is that sum of voltage drop in a closed loop is equal to zero. Inductors have a unique property by which they oppose the change in magnetic flux linked to them. The voltage drop across resistor is V_R = IR and across inductor is L \( \frac{dI}{dt} \). In the steady state current through inductor becomes constant which leads to zero voltage drop across inductor. ie. it behaves like short circuit. Refer to circuit given in the figure. E = 10 V, R₁ = 2Ω, R₂ = 3Ω, R₃ = 6Ω and L = 5H.
Now answer the following questions

[Figure: A circuit with a switch S, a cell E and a resistor R₁ in series carrying current I₁; at a junction the current divides into I₂ through R₂ in series with an inductor L, and I₃ through R₃ connected across the junction and the return wire]

4. The current I₁ just after pressing the switch S is
(a) \( \frac{10}{8} \) A
(b) \( \frac{10}{5} \) A
(c) \( \frac{10}{12} \) A
(d) \( \frac{10}{6} \) A

Answer: (a) \( \frac{10}{8} \) A

1. Just after the switch is closed, the inductor opposes any change of current, so I₂ = 0 and the branch with R₂ and L carries no current.
2. The current flows through R₁ and R₃ only: I₁ = \( \frac{10}{2 + 6} \) = \( \frac{10}{8} \) A.
3. Hence, the correct option is (a).

Teacher's Note:
a) At the first instant an inductor behaves like an open circuit.
b) Do not use the steady-state (short-circuit) idea for the instant just after closing the switch.

5. The current I₁ long after pressing the switch S is
(a) \( \frac{10}{4} \) A
(b) \( \frac{10}{5} \) A
(c) \( \frac{10}{12} \) A
(d) \( \frac{10}{6} \) A

Answer: (a) \( \frac{10}{4} \) A

1. Long after closing the switch, the inductor behaves like a plain wire, so R₂ and R₃ are in parallel.
2. R₂₃ = \( \frac{3 \times 6}{3 + 6} \) = 2 Ω.
3. I₁ = \( \frac{10}{2 + 2} \) = \( \frac{10}{4} \) A.
4. Hence, the correct option is (a).

Teacher's Note:
a) In the steady state the voltage drop across an ideal inductor is zero.
b) Find the parallel combination first and then add R₁ in series.

6. The current I₂ long after pressing the switch S is
(a) \( \frac{10}{4} \) A
(b) \( \frac{10}{5} \) A
(c) \( \frac{10}{12} \) A
(d) \( \frac{10}{6} \) A

Answer: (d) \( \frac{10}{6} \) A

1. I₁ = \( \frac{10}{4} \) A divides between R₂ (3 Ω) and R₃ (6 Ω) in the inverse ratio of resistances.
2. I₂ = \( \frac{10}{4} \) × \( \frac{6}{3 + 6} \) = \( \frac{10}{6} \) A.
3. Hence, the correct option is (d).

Teacher's Note:
a) In a current divider the smaller resistance takes the larger share of the current.
b) Check: I₃ = \( \frac{10}{12} \) A and I₂ + I₃ = \( \frac{10}{4} \) A.

7. The current through R₂ just after releasing the switch S is
(a) \( \frac{10}{4} \) A
(b) \( \frac{10}{5} \) A
(c) \( \frac{10}{6} \) A
(d) \( \frac{10}{12} \) A

Answer: (c) \( \frac{10}{6} \) A

1. Before the switch is released, the current in the inductor is I₂ = \( \frac{10}{6} \) A.
2. The inductor current cannot change suddenly, so it continues to flow at \( \frac{10}{6} \) A through the closed loop of L, R₂ and R₃.
3. This loop current passes through R₂, so the current through R₂ is \( \frac{10}{6} \) A.
4. Hence, the correct option is (c).

Teacher's Note:
a) An inductor keeps its current unchanged at the instant the circuit is broken.
b) The direction of the current in R₃ reverses, but the current in R₂ keeps its value and direction.

Each of the questions given below consists of two statements, an assertion (A) and reason (R). Select the number corresponding to the appropriate alternative as follows
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

8. A: We prefer a potentiometer with a longer bridge wire.
R: By doing this, the sensitivity of the bridge is increased because for longer wire potential drop per unit length will be small.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (a) If both A and R are true and R is the correct explanation of A

1. A longer potentiometer wire has a smaller potential gradient (potential drop per unit length).
2. A small potential gradient means a balance point can be located to a finer precision, so the sensitivity increases.
3. Hence, both A and R are true and R explains A, so the correct option is (a).

Teacher's Note:
a) Sensitivity of a potentiometer rises when the potential gradient falls.
b) A longer wire also reduces the percentage error in the length reading.

9. A: In steady state the branch of the circuit containing a capacitor acts as a open circuit.
R: In steady state, inductor acts as a conductor of zero resistance.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (b) If both A and R are true but R is not the correct explanation of A

1. In the steady state a capacitor is fully charged and carries no current, so its branch acts as an open circuit. A is true.
2. In the steady state the current in an inductor is constant, so the voltage across it is zero and it acts as a conductor of zero resistance. R is true.
3. R describes the inductor and does not explain the behaviour of the capacitor, so the correct option is (b).

Teacher's Note:
a) A capacitor blocks steady current; an inductor passes it freely.
b) Both statements are true, but the reason is about a different element, so it is not the explanation.

10. A: Ohm’s law and Kirchhoff’s laws are contradictory.
R: Electrons in a conductor have no motion in the absence of a potential difference across it.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: Both A and R are false

1. Ohm's law (V = IR) and Kirchhoff's laws are consistent with each other; Kirchhoff's laws are used together with Ohm's law to solve circuits. A is false.
2. In the absence of a potential difference, free electrons still move randomly with thermal speeds; only their net drift is zero. R, as stated, is false.
3. Answer as per the printed key: (d) If A is false but R is true.

Teacher's Note:
a) Kirchhoff's laws come from conservation of charge and energy and fit with Ohm's law.
b) Zero drift velocity is not the same as no motion at all.

11. A: Current is a vector quantity.
R: It obeys vector addition.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: Both A and R are false

1. Current is a scalar quantity although it has a direction along the conductor. A is false.
2. Currents are added algebraically at a junction, not by the vector law. R is false.
3. None of the four printed options says that both statements are false; the printed key shows (d).

Teacher's Note:
a) Having a direction does not make a quantity a vector; it must also obey the vector law of addition.
b) Kirchhoff's junction rule adds currents as ordinary numbers.

Answer as per the printed key: (d) If A is false but R is true.

12. A: The wire of a potentiometer should be of uniform area of cross section.
R: The resistance per unit length and hence fall of potential per unit of length in case of a uniform wire is constant.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (a) If both A and R are true and R is the correct explanation of A

1. For a wire of uniform cross section and material, resistance per unit length is constant.
2. A constant current then gives a uniform fall of potential per unit length, so the potential is proportional to length.
3. This is exactly why the wire must be uniform, so the correct option is (a).

Teacher's Note:
a) The principle V proportional to l holds only if the potential gradient is uniform.
b) A non-uniform wire would make the balance length meaningless for comparing e.m.f.s.

13. A: Fuse wire should have high resistance but low melting point.
R: It is used to control the current.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (c) If A is true but R is false

1. A fuse wire is made of a material with a low melting point and fairly high resistance, so that it heats up quickly and melts when the current exceeds the safe value. A is true.
2. A fuse does not control the current; it protects the circuit by breaking it when the current is too large. R is false.
3. Hence, the correct option is (c).

Teacher's Note:
a) A fuse is a safety device that breaks the circuit; it does not regulate the current.
b) The printed key shows (d), but R is not correct, so (c) is the correct option.

14. A: Terminal voltage of a cell is greater than e.m.f. of the cell, during charging of the cell.
R: For charging, current must flow from positively to the negative plate inside the cell.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (a) If both A and R are true and R is the correct explanation of A

1. During charging, current enters the positive terminal and passes through the cell from the positive plate to the negative plate. R is true.
2. The terminal voltage is then V = E + Ir, which is greater than E. A is true.
3. The direction of current in R is what gives V = E + Ir, so R explains A and the correct option is (a).

Teacher's Note:
a) Discharging gives V = E - Ir, while charging gives V = E + Ir.
b) Remember that the external source must have a higher e.m.f. than the cell being charged.

15. A: The element of a heater is very hot, while the conducting wire carrying the current are cold.
R: It happens because the resistance of connecting wires is high as compared to that of element of heater.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (c) If A is true but R is false

1. The same current flows through the element and the connecting wires, and the heat produced is I²R t.
2. The element has a much higher resistance than the connecting wires, so it becomes very hot while the wires stay cold. A is true.
3. R says the resistance of the connecting wires is higher, which is false.
4. Hence, the correct option is (c).

Teacher's Note:
a) Heating in a series circuit is proportional to resistance because the current is the same.
b) The printed key shows (a), but R states the opposite of the reason, so (c) is correct.
c) Connecting wires are made of copper to keep their resistance low.

16. A: Higher the range greater is the resistance of ammeter.
R: To increase the range of ammeter shunt has to be increased.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: Both A and R are false

1. A larger range needs a smaller shunt resistance, so the combined resistance of the ammeter becomes smaller as the range increases. A is false.
2. To increase the range, the shunt resistance has to be decreased, not increased. R is false.
3. None of the four printed options says that both statements are false; the printed key shows (d).

Teacher's Note:
a) An ideal ammeter has zero resistance; a high-range ammeter is closer to ideal.
b) Range I = I_g (1 + G/S), so range rises when S falls.

17. A: The range of an ammeter can be decreased.
R: The value of shunt, S = \( \frac{I_g G}{I - I_g} \). S can not be negative. Therefore range of ammeter can not be decreased.
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true

Answer: (c) If A is true but R is false

1. The range of an ammeter can be reduced by increasing the shunt resistance, down to the value I_g of the galvanometer. A is true.
2. S cannot be negative, so I must be greater than I_g; this means the range cannot be made less than I_g, but it can still be decreased.
3. The conclusion in R is false, so the correct option is (c).

Teacher's Note:
a) The range of an ammeter made from a galvanometer can never be less than I_g.
b) Do not read "cannot be negative" as "cannot be decreased".