RBSE Class 12 Physics Chapter 7 — Alternating Current: Most Repeated Questions (2013–2026)

Φ=NBAcos(ωt)

ε=-dΦ/dt

ε=-d/dt(NBAcos(ωt)

ε=-NBAd/dt(cos (ωt))

ε=-NBA(-ωsin(ωt))

ε=NBAωsin (ωt)

sin(ωt)=1

ε₀=NBAω

ε=ε₀sin(ωt)

All solved questions for this chapter are in the full RBSE Physics PYQ book.

(A)

P=v×I →P=(V₀sinωt)×(I₀sin(ωt-π/2)); →P=V₀I₀sinωt×(-cosωt); →P=-V₀I₀/2(2sinωtcosωt); →P=-V₀I₀/2sin2ωt

Integrate both sides (Power Average P_avg =1/T∫₀^TPdt )

∫Pdt=-V₀I₀/2∫₀^Tsin2ωtdt

→∫Pdt=-V₀I₀/2[-cos2ωt/2ω]₀^T; →∫Pdt=+V₀I₀/4ω[cos2ωT-cos0]

Since ωT=2π

→∫Pdt=V₀I₀/4ω[cos4π-cos0]; →∫Pdt=V₀I₀/4ω[1-1]; →∫Pdt=0; ⇒[P_avg =0]

(B) Z=√(R²+(X_L-X_C)²)

Where:

R is the resistance (24Ω)

X_L is the inductive reactance (110Ω)

X_C is the capacitive reactance (110Ω)

Substituting the given values:

Z=√((24Ω)²+(110Ω-110Ω)²); Z=√((24Ω)²+(0Ω)²); Z=√((24Ω)²); Z=24Ω

All solved questions for this chapter are in the full RBSE Physics PYQ book.

For a series LCR circuit with applied voltage V = V₀sinωt, the same current I = I₀sin(ωt−φ) flows through all three elements

The voltage phasors are: Vᵣ (along I), Vₗ (leading I by 90°), Vᴄ (lagging I by 90°)

Resultant voltage (phasor sum, with Vₗ and Vᴄ opposing each other along the same line, perpendicular to Vᵣ):

V₀=√(V_R²+(V_L-V_C)²)

Dividing through by I₀

Vᵣ=I₀R,

Vₗ=I₀Xₗ,

Vᴄ=I₀Xᴄ

Z=V₀/I₀=√(R²+(X_L-X_C)²)

where X_L=ω L and X_C=1/ω C

All solved questions for this chapter are in the full RBSE Physics PYQ book.

Resonance is the phenomenon where a system oscillates with maximum amplitude when the frequency of an applied external periodic force matches the system's own natural frequency

Any Series LCR is said to be in resonance when ' Z ' of circuit is minimum (' I ' is max ​^m ). Which means, X_L=X_C

Frequency of LCR at resonance: As, X_L=X_C

ωL=1/ωC

ω²=1/LC

ω_r=1/√(LC) or, 2πf_r=1/√(LC); f_r=1/2π√(LC)

The graph peaks where the external frequency ( ω ) equals the resonant frequency ( ω₀ ), at which point the circuit's opposition (Impedance) is minimum (Z=R) and the Current (I_m) is maximum

​The sharpness of this peak is the circuit's selectivity (or Q-factor), which is inversely proportional to resistance (R): lower R gives a taller, sharper peak

All solved questions for this chapter are in the full RBSE Physics PYQ book.

1. Copper Loss =

Heat generated in windings due to resistance

Minimized by using thick, low-resistance copper wires

2. Eddy Current Loss:

Heating occurs in the core due to induced currents

Minimized by using a laminated iron core

Electric power is transmitted at high voltage over long distances to minimize loss in the transmission cables

P loss = I² R

A lower current result in much less power loss hence, electric power is transmitted at high voltage over long over long distance

All solved questions for this chapter are in the full RBSE Physics PYQ book.