JEE Mains Physics · Alternating Current

LC Oscillations, Transformers and AC Devices

A charged capacitor across an inductor swaps its energy back and forth at ω = 1/√(LC); a transformer changes a.c. voltage in the ratio of its turns while the power, less its losses, carries through.

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Why this matters

Fifteen PYQs, eleven of them multiple choice. Five are LC oscillations: the largest current, a new frequency, or how long the energy takes to move into the inductor. Ten are transformers and devices: turns, voltages and currents with or without losses, and match lists of the principle each device works on. Energy conservation and the turns ratio do almost all the work.

Concept 1 of 2: LC oscillations

Connect a charged capacitor across an inductor and the charge flows out as a current. The inductor keeps the current going after the capacitor is empty, charging it the other way. Energy moves from the capacitor's electric field to the inductor's magnetic field and back, like a mass on a spring. With no resistance the total never changes, so the largest current comes from setting all the energy in the inductor.

Definition

Starting with charge $Q_0$ : $q = Q_0\cos\omega t$ , $i = -Q_0\omega\sin\omega t$ , with $\omega = \dfrac{1}{\sqrt{LC}}$ .
Energy: $U_C = \dfrac{q^{2}}{2C}$ , $U_L = \dfrac{1}{2}Li^{2}$ , and $U_C + U_L = \dfrac{Q_0^{2}}{2C}$ at every instant.
Largest current: $I_{max} = Q_0\omega = \dfrac{Q_0}{\sqrt{LC}} = V_0\sqrt{\dfrac{C}{L}}$ , from $\dfrac{1}{2}CV_0^{2} = \dfrac{1}{2}LI_{max}^{2}$ .
Energy share to time: if the inductor holds a fraction k of the energy, the capacitor holds $1 - k$ , so $q = Q_0\sqrt{1 - k}$ . Solve $\cos\omega t = \sqrt{1 - k}$ .
Scaling: $\omega \propto \dfrac{1}{\sqrt{LC}}$ . L times a and C times b give $\dfrac{\omega}{\sqrt{ab}}$ .

LC oscillator

\omega = \frac{1}{\sqrt{LC}} \qquad I_{max} = \frac{Q_0}{\sqrt{LC}} = V_0\sqrt{\frac{C}{L}}

Worked example

5\ \mu\text{F}

capacitor charged to 40 V is connected across a 0.2 H inductor. Find

\omega

, the largest charge and the largest current.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q18Moderate

Example 1 · Alternating Current · LC Oscillations, Transformers and AC Devices

A capacitor of capacitance

100\mu F

is charged to a potential of

12\text{ }V

and connected to a

6.4mH

inductor to produce oscillations. The maximum current in the circuit would be:

√(C/L) or √(L/C)

From ½CV₀² = ½LI², the current is V₀√(C/L). A large capacitor or a small inductor gives a large current.

Charge fraction and energy fraction

Energy goes as q². Half the charge leaves a quarter of the energy in the capacitor, not half.

Drill 4 more on lc oscillations

Concept 2 of 2: Transformers and a.c. devices

Both coils of a transformer share one changing flux, so each turn has the same emf. Voltage therefore goes in the ratio of the turns. Power cannot be made, so where the voltage goes up the current comes down. A real transformer loses a little, and its efficiency says how much of the input power comes out.

Definition

$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ . Step-up if $N_s > N_p$ , step-down if $N_s < N_p$ .
Ideal: $V_pI_p = V_sI_s$ , so $\dfrac{I_s}{I_p} = \dfrac{N_p}{N_s}$ . The current changes the opposite way to the voltage.
Efficiency: $\eta = \dfrac{P_{out}}{P_{in}}$ , so $V_sI_s = \eta V_pI_p$ .
A resistive load R on the secondary: $P = \dfrac{V_s^{2}}{R}$ . Seen from the primary it looks like $\left(\dfrac{N_p}{N_s}\right)^{2}R$ .
A.c. generator: electromagnetic induction, turning mechanical energy into electrical.
Transformer: mutual induction between two coils. Galvanometer: detects a current. Quality factor: the sharpness of resonance. Resonance needs both L and C. A metal detector is described in NCERT as using resonance in an a.c. circuit.

Transformer

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

Worked example

A transformer steps 11 kV down to 440 V. Its primary has 2500 turns and takes 4 A, and its efficiency is

88\%

. Find the number of secondary turns and the secondary current.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 30 Jan 2024 · Q111Moderate

Example 2 · Alternating Current · LC Oscillations, Transformers and AC Devices

A power transmission line feeds input power at 2.3

kV

to a step -down transformer with its primary winding having 3000 turns. The output power is delivered at

230\text{ }V

by the transformer. The current in the primary of the transformer is

5\text{ }A

and its efficiency is

90\%

. The winding of transformer is made of copper. The output current of transformer is _____ A.

Turning the ratio upside down

Voltage follows the turns: more turns, more volts. Current goes the other way. Write

\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}

before putting numbers in.

Forgetting the efficiency on the current

With losses, the output power is η times the input. Find the output current from that power, not from the ideal turns ratio.

Drill 9 more on transformers and a.c. devices

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 (2)

LC oscillations
LC oscillator
$\omega = \frac{1}{\sqrt{LC}} \qquad I_{max} = \frac{Q_0}{\sqrt{LC}} = V_0\sqrt{\frac{C}{L}}$
Transformers and a.c. devices
Transformer
$\frac{V_s}{V_p} = \frac{N_s}{N_p} \qquad V_sI_s = \eta\,V_pI_p$

Watch out for (4)

√(C/L) or √(L/C)
→ LC oscillations
Charge fraction and energy fraction
→ LC oscillations
Turning the ratio upside down
→ Transformers and a.c. devices
Forgetting the efficiency on the current
→ Transformers and a.c. devices

Test yourself on Alternating Current

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