CBSE Class 12 Physics Alternating Current Important Questions

CBSE Class 12 Physics Alternating Current 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.

IMPORTANT ASSIGNMENT
Comprehension

The a.c. generator which is one of the most important applications of the phenomenon of electromagnetic induction converts mechanical energy into electrical energy. A rectangular coil consisting of a large number of turns of copper wire would over a soft iron core is rotated between the pole pieces of a permanent strong magnet. The magnetic flux through the coil changes continuously with time, thus producing induced emf given by E = E₀ sin ωt
When a load resistor R is connected across the terminals, a current I flows through the circuit.
I = \( \frac{E}{R} \) = \( \frac{E_0}{R} \) sin ωt = I₀ sin ωt
Where I₀ = E₀/R. Such a current is called a.c. or alternating current.

1. In an a.c. generator, a coil of area A and having N turns rotates in a magnetic field B. The magnetic flux through the coil is
(a) maximum equal to NAB when the plane of the coil is perpendicular to the magnetic field.
(b) zero when the plane of the coil is parallel to the field.
(c) 1/2 NAB when the plane of the coil makes an angle of 60° with the field.
(d) 1/4 NAB when the plane of the coil makes an angle of 30° with the field.

Answer: (a) and (b)

1. Flux = NAB cos θ, where θ is the angle between the normal to the coil and the field.
2. Plane perpendicular to the field: θ = 0°, so flux = NAB, which is the maximum. Statement (a) is true.
3. Plane parallel to the field: θ = 90°, so flux = 0. Statement (b) is true.
4. Plane at 60° with the field: θ = 30°, so flux = NAB cos 30° = 0.87 NAB, not 1/2 NAB. Statement (c) is false.
5. Plane at 30° with the field: θ = 60°, so flux = NAB cos 60° = 1/2 NAB, not 1/4 NAB. Statement (d) is false.

Teacher's Note:
a) Flux uses the angle between the field and the normal to the coil, not the plane.
b) The printed key gives (a), (b), (c); statement (c) is true only if 60° is measured from the normal, so the correct choice for the wording as printed is (a) and (b).
c) Quick check: the plane angle and the normal angle add up to 90°, so cos and sin swap.

2. In an a.c. generator, initially (i.e. at t = 0) the plane of the coil is normal to the magnetic field. Which graph shown in the figure represents the variation of induced emf E with time t ?

[Figure: Four graphs of E against t. (a) a single positive hump starting near zero; (b) a single negative dip that returns towards zero; (c) a sine wave starting from zero, rising first and then going negative; (d) a constant horizontal line above the t-axis.]

Answer: (c)

1. At t = 0 the plane of the coil is normal to the field, so the flux is maximum and the rate of change of flux is zero.
2. Induced emf E = -dφ/dt = NABω sin ωt, which is zero at t = 0.
3. The emf then rises sinusoidally, changes sign every half cycle and repeats.
4. This is the sine wave starting from zero, graph (c).

Teacher's Note:
a) Maximum flux means zero emf; emf is the slope of the flux curve.
b) Graph (d) is d.c. and graphs (a) and (b) are single pulses, none of which is a continuous a.c.

3. In an a.c. generator, the peak value of the induced emf depends upon the
(a) frequency of rotation of the coil
(b) area of the coil
(c) number of turns in the coil
(d) strength of the magnetic field

Answer: (a), (b), (c) and (d)

1. Induced emf E = NABω sin ωt, so the peak value is E₀ = NABω.
2. Here ω = 2πf, so E₀ = 2πf NAB.
3. E₀ depends on f, A, N and B, so all four options are correct.

Teacher's Note:
a) Peak emf is proportional to N, A, B and the angular speed.

4. In an a.c. generator
(a) the coil is wound over a soft iron core in order to increase the flux.
(b) an electromagnetic field with an alternating current in used.
(c) the output is always taken across a load resistor
(d) the mechanical energy of the rotating coil is converted into electrical energy

Answer: (a), (c) and (d)

1. Soft iron has high permeability, so winding the coil over it increases the flux. Statement (a) is true.
2. A generator uses a strong permanent magnet (or a d.c. electromagnet), not an alternating field. Statement (b) is false.
3. The load resistor R is connected across the terminals of the coil to take the output. Statement (c) is true.
4. The generator converts the mechanical energy of rotation into electrical energy. Statement (d) is true.

Teacher's Note:
a) The field is steady; only the flux through the rotating coil changes.
b) Energy flow is mechanical to electrical, the reverse of a motor.

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

5. A: A d.c. voltmeter can not read alternate current .
R: It is because, average value of a.c. is zero over a half cycle.

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

1. A d.c. voltmeter reads the average value of the current, and the average of a.c. over a full cycle is zero, so it shows no steady reading. A is true.
2. Over a half cycle the average value is not zero; it equals 2I₀/π = 0.637 I₀. R is false.
3. Hence, the correct option is (c).

Teacher's Note:
a) The average of a.c. is zero over a full cycle, not over a half cycle.
b) The printed key gives (d), but A is true, so (c) is correct.
c) Hot-wire and moving-iron meters read a.c. because they respond to the heating effect, which does not average to zero.

6. A: The maximum value of power factor of a.c. circuit is 1 and minimum value is -1.
R: The power consumed in a purely capacitive circuit is maximum.

Answer: Both A and R are false, so none of the options (a) to (d) applies.

1. Power factor = cos φ = R/Z, and R and Z are both positive in a passive circuit, so it lies between 0 and 1. The minimum is 0, not -1. A is false.
2. In a purely capacitive circuit φ = 90°, so power = VI cos 90° = 0. The power consumed is zero, not maximum. R is false.
3. Since both statements are false, the correct reading is that both are false.

Teacher's Note:
a) A pure capacitor or inductor is a wattless circuit: the average power is zero.
b) The printed key (d) treats R as true, which is wrong; the power factor of such a circuit is 0, which is its minimum value.
c) Power factor is never negative for a passive RLC circuit.

7. A: In a series RCL circuit the voltages across resistor, inductor and capacitor are 8V, 16 V and 10 V respectively. The resultant e.m.f. in the circuit is 10 V.
R: E = \( \sqrt{V_R^2 + (V_L - V_C)^2} \) .

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

1. For a series RLC circuit, E = \( \sqrt{V_R^2 + (V_L - V_C)^2} \). So R is true.
2. E = \( \sqrt{8^2 + (16 - 10)^2} \) = \( \sqrt{64 + 36} \) = \( \sqrt{100} \) = 10 V. So A is true.
3. A is obtained directly by using R, so R explains A.
4. Hence, the correct option is (a).

Teacher's Note:
a) The voltages across L and C are in opposite phase, so they are subtracted before squaring.
b) Adding 8 + 16 + 10 directly is the common mistake.

8. A. In LCR parallel resonant circuit, impedance of the circuit is maximum.
R. In LCR series resonant circuit, impedance of the circuit is minimum.

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

1. At resonance in a parallel LCR circuit the net current drawn is minimum, so the impedance is maximum. A is true.
2. At resonance in a series LCR circuit X_L = X_C, so Z = R, which is the minimum. R is true.
3. R describes a different circuit, so it does not explain A.
4. Hence, the correct option is (b).

Teacher's Note:
a) Series resonance is an acceptor circuit with minimum Z; parallel resonance is a rejector circuit with maximum Z.
b) Do not treat the two statements as cause and effect just because both mention resonance.

9. A. Space offers impedance to the motion of e.m. waves through it and this impedance limits the velocity of e.m. waves through vacuum.
R. The impedance of free space to the flow of e.m. waves is 120 π Ω.

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

1. Free space has an impedance Z₀ = \( \sqrt{\frac{\mu_0}{\varepsilon_0}} \), which is related to the speed of e.m. waves c = \( \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \). A is taken as true.
2. Z₀ = \( \sqrt{\frac{4\pi \times 10^{-7}}{8.85 \times 10^{-12}}} \) ≈ 377 Ω = 120π Ω. R is true.
3. The value 120π Ω only states the magnitude; it does not explain why the velocity is limited.
4. Hence, the correct option is (b).

Teacher's Note:
a) 120π Ω is about 377 Ω, the characteristic impedance of free space.
b) A number given as a reason cannot explain a physical statement.

10. A. In a resonant circuit, reactance of the LCR circuit is zero.
R. Resonant frequency of LCR circuit is independent of ohmic resistance of the circuit.

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

1. Reactance X = X_L - X_C, which becomes zero when ωL = \( \frac{1}{\omega C} \).
2. This gives the resonant frequency ω₀ = \( \frac{1}{\sqrt{LC}} \), which contains only L and C and not R. So R is true.
3. Zero reactance happens exactly at this frequency, so R explains A.
4. Hence, the correct option is (a).

Teacher's Note:
a) Resistance changes the sharpness of resonance (Q factor), not its frequency.
b) At resonance the circuit is purely resistive, Z = R.