MHT-CET Physics · AC Circuits

LC Oscillations, the Transformer and the AC Generator

A charged capacitor discharging through an inductor swaps its energy back and forth at ω = 1/√(LC); a transformer steps voltage by the turns ratio; and a coil turning in a field generates e₀ = NABω.

All AC Circuits notes MHT-CET Physics strategy

Why this matters

9 PYQs, three HARD. Three separate ideas share this page: the energy swap of an LC circuit (its peak current and the quarter-period timing), the transformer's turns and efficiency, and the peak e.m.f. and power of a rotating coil.

Concept 1 of 3: LC Oscillations

With no resistance, the energy that sat in the capacitor's field moves into the inductor's field and back, over and over. All of ½CV² becomes ½LI₀² at the moment the capacitor is empty — which is a quarter-period after it was full.

Definition

$\omega = \dfrac{1}{\sqrt{LC}}$ , $T = 2\pi\sqrt{LC}$ .
Energy: $\dfrac{1}{2}CV^2 = \dfrac{1}{2}LI_0^2 \Rightarrow I_0 = V\sqrt{\dfrac{C}{L}}$ . The instantaneous current can take any value up to this.
Capacitor full to current greatest: $\dfrac{T}{4} = \dfrac{\pi}{2}\sqrt{LC}$ . The current keeps reversing; the energy is all magnetic only at the instants the capacitor is empty.
Inductor on a.c.: its stored energy goes from greatest to zero in $\dfrac{T}{4}$ .

LC circuit

T = 2\pi\sqrt{LC}, \qquad I_0 = V\sqrt{\frac{C}{L}}

Worked example

4\,\mu

F capacitor at 100 V discharges through a 1 mH inductor. Peak current and period?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q25Moderate

Example 1 · AC Circuits · LC Oscillations, Transformer, and AC Generator

1\ \mu\text{F}

capacitor is charged to 50 V and is then discharged through 10 mH inductor of negligible resistance. The maximum current in the inductor is

Taking the full period for 'current greatest'

From a full capacitor, the current is greatest a QUARTER period later.

2\pi\sqrt{LC}

is the time to come back to a full capacitor of the same polarity.

Drill 4 more on lc oscillations

Concept 2 of 3: The Transformer

A transformer passes power from one coil to another through a shared changing flux, so the voltage per turn is the same in both. An ideal one loses nothing: it raises voltage only by lowering current.

Definition

$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ ; ideal: $V_pI_p = V_sI_s$ , so $\dfrac{I_s}{I_p} = \dfrac{N_p}{N_s}$ .
Efficiency $\eta = \dfrac{V_sI_s}{V_pI_p}$ : 3 kW in at 200 V and 90% ⇒ $I_p = 15$ A and 2700 W out.
A transformer works only on a.c.

Transformer

\frac{V_s}{V_p} = \frac{N_s}{N_p}, \qquad \eta = \frac{V_sI_s}{V_pI_p}

Worked example

An ideal transformer steps 200 V up to 800 V. The primary has 100 turns; the secondary delivers 2 A. Secondary turns and primary current?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q34Moderate

Example 2 · AC Circuits · LC Oscillations, Transformer, and AC Generator

A transformer having efficiency 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil are respectively

Applying the efficiency to the primary current

The primary current comes from the INPUT power,

\frac{3000}{200} = 15

A. The 90% applies to what comes out:

V_s = \frac{2700}{6} = 450

Drill 1 more on the transformer

Concept 3 of 3: The AC Generator

A coil turning in a magnetic field has its flux change sinusoidally, so it produces e = NABω sin ωt. The peak e.m.f. grows with every factor — turns, area, field and speed.

Definition

$e = NAB\omega\sin\omega t$ , $e_0 = NAB\omega$ .
Into a resistance $R$ : average power $\dfrac{e_0^2}{2R} = \dfrac{N^2A^2B^2\omega^2}{2R}$ .

Generator e.m.f.

e_0 = NAB\omega, \qquad \bar P = \frac{(NAB\omega)^2}{2R}

Worked example

A 50-turn coil of area 0.02 m² spins at 100 rad/s in a 0.5 T field. Peak e.m.f.?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q1Moderate

Example 3 · AC Circuits · LC Oscillations, Transformer, and AC Generator

A circular coil of resistance

R

, area

A

, number of turns

N

is rotated about its vertical diameter with angular speed

\omega

in a uniform magnetic field of magnitude

B

. The average power dissipated in a complete cycle is

Squaring only part of e₀

Power goes as

e_0^2

, so as

N^2A^2B^2\omega^2

— every factor squared. Options with

NAB\omega

to the first power are the e.m.f., not the power.

Drill 1 more on the ac generator

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

LC Oscillations
LC circuit
$T = 2\pi\sqrt{LC}, \qquad I_0 = V\sqrt{\frac{C}{L}}$
The Transformer
Transformer
$\frac{V_s}{V_p} = \frac{N_s}{N_p}, \qquad \eta = \frac{V_sI_s}{V_pI_p}$
The AC Generator
Generator e.m.f.
$e_0 = NAB\omega, \qquad \bar P = \frac{(NAB\omega)^2}{2R}$

Watch out for (3)

Taking the full period for 'current greatest'
→ LC Oscillations
Applying the efficiency to the primary current
→ The Transformer
Squaring only part of e₀
→ The AC Generator

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