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Primary Current of an Ideal Transformer

An ideal transformer has a turns ratio $N_p : N_s = 10 : 1$ and a primary voltage of $2400\ \text{V}$ .

The secondary supplies a purely resistive load of $5.0\ \Omega$ .

Find the primary current.

Given

a	10 -	Turns ratio Np/Ns
Vp	2400 V	Primary voltage
RL	5 ohm	Secondary load resistance

Hint 1

Work through the secondary side first: voltage, then current, then reflect back.

Hint 2

Voltage steps down by a, so current must step up by a for power to balance.

Hint 3

I_p = I_s/a. Check by confirming V_pI_p = V_sI_s.

Worked solution — try the problem first

Step 1 — Secondary voltage. Voltage scales down with the turns ratio:

$V_s = \frac{V_p}{a} = \frac{2400}{10} = 240\ \text{V}$

Step 2 — Secondary current from the load:

$I_s = \frac{V_s}{R_L} = \frac{240}{5.0} = 48\ \text{A}$

Step 3 — Primary current. Current scales inversely to voltage — a step-down transformer steps current up on the secondary side:

$I_p = \frac{I_s}{a} = \frac{48}{10} = 4.8\ \text{A}$

$\boxed{I_p = 4.8\ \text{A}}$

Power check. An ideal transformer conserves power:

$P_p = 2400 \times 4.8 = 11{,}520\ \text{W}, \qquad P_s = 240 \times 48 = 11{,}520\ \text{W} \ \checkmark$

Shortcut via reflected impedance. The load appears from the primary as $a^2R_L = 100 \times 5 = 500\ \Omega$ , so $I_p = 2400/500 = 4.8$ A directly. Note impedance transforms with the square of the turns ratio.

Concepts:Ideal transformerTurns ratioReflected impedancePower conservation

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