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**Q.1. A series LCR circuit is connected to an AC voltage source. When L is removed from the circuit, the phase difference between current and voltage is π/3. If instead C is removed from the circuit, the phase difference is again π/3 between current and voltage. The power factor of the circuit is : (2020)A: 1.0 B: –1.0 C: zero D: 0.5Ans:** A

When inductor alone is removed,

When capacitor alone is removed,

Thus, for the original circuit,

power factor = cos Δϕ = R/Z = R/R = 1

A: Conductor

B: Inductor

C: Switch

D: Fuse

Ans:

Solution:

Fuse wire has less melting point so when excess current flows, due to heat produced in it, it melts.

A: Zero, 60 μA

B: 60 μA, 60 μA

C: 60 μA, zero

D: Zero, zero

Ans:

Solution:

Capacitance of capacitor C = 20 mF

= 20 × 10

Rate of change of potential

i

= 60 × 10

= 60 μA

As we know that i

A: 0.138 H

B: 138.88 H

C: 1.389 H

D: 13.89 H

Ans:

Solution:

Energy stored in the inductor

Energy stored in the inductor

A: I

B: I

C: I

D: I

Ans:

Solution:

**Q.6. ****An inductor 20 mH, a capacitor 100 μF and a resistor 50 Ω are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is (2018)A: 0.79 WB: 0.43 WC: 2.74 WD: 1.13 WAns: **A

Solution:

**A: 0.2 AB: 2 AC: 0 ampereD: 2 mAAns:** B

Solution:

**Q.8. ****An inductor 20 mH, a capacitor 50 μF and a resistor. 40Ω are connected in series across a source of emf V = 10 sin 340 t. The power loss in A.C. circuit is (2016)A: 0.89WB: 0.51WC: 0.67WD: 0.76WAns: **B

Solution:

**Q.9. ****A small signal voltage V(t) = V _{0} sinωt is applied across an ideal capacitor C : (2016)A: Current I(t), leads voltage V(t) by 180 degreeB: Current I(t), lags voltage V(t) by 90 degreeC: Over a full cycle the capacitor C does not consume any energy from the voltage source.D: Current I(t) is in phase with voltage V(t)Ans: **C

Solution:

Capacitor does not consume energy effectively over full cycles

A: P

B:

C:

D:

Ans: B

Solution:

A resistance R draws power P when connected to an AC source.

The magnitude of voltage of the AC source is

V^{2} = RP

∴ V=

An inductor of inductance L and reactance ωL is now placed in series with the resistance

**Q.11. ****A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6A, the voltage across the secondary coil and the current in the primary coil respectively are : (2014)A: 450 V, 13.5 AB: 600 V, 15 AC: 300 V, 15 AD: 450 V, 15 A**

Initial power = 3000 W

As the efficiency is 90%, then final power

That is,

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