Vs/Vp = Ns/Np = Ip/Is. Efficiency and energy losses.

Transformer — Step Up and Step Down

Question

A transformer has 500 turns in its primary coil and 2000 turns in its secondary coil. If the primary voltage is 220 V and the primary current is 4 A, find: (a) the secondary voltage, (b) the secondary current, assuming 100% efficiency.

Solution — Step by Step

The secondary has more turns than the primary (2000 > 500), so this is a step-up transformer — voltage increases from primary to secondary. This orientation is used in long-distance power transmission.

The fundamental transformer equation relates voltages to the turns ratio:

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

Plugging in:

\frac{V_s}{220} = \frac{2000}{500} = 4

V_s = 220 \times 4 = \textbf{880 V}

At 100% efficiency, input power equals output power:

V_p \cdot I_p = V_s \cdot I_s

220 \times 4 = 880 \times I_s

I_s = \frac{880}{880} = \textbf{1 A}

For an ideal transformer, the current ratio is the inverse of the turns ratio:

\frac{I_s}{I_p} = \frac{N_p}{N_s} = \frac{500}{2000} = \frac{1}{4}

I_s = \frac{4}{4} = 1 \text{ A} \checkmark

Both approaches agree — good sign that we haven’t made an error.

Why This Works

The transformer works on the principle of mutual induction. An alternating current in the primary coil creates a changing magnetic flux in the iron core. This changing flux threads through the secondary coil and induces an EMF proportional to the number of turns.

More turns in the secondary means more “cuts” of the same changing flux — so each additional turn adds to the total induced EMF. This is why $V \propto N$ holds so cleanly.

The current relationship follows directly from energy conservation. A step-up transformer increases voltage but must decrease current to keep power constant — you can’t get more energy out than you put in. This is why high-voltage transmission lines carry lower current, reducing heat loss ( $P = I^2 R$ ) over long distances.

\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}

Efficiency:

\eta = \frac{V_s \cdot I_s}{V_p \cdot I_p} \times 100\%

For ideal transformer: $\eta = 100\%$ , so $V_p I_p = V_s I_s$

Alternative Method

We can skip finding $V_s$ first and directly use the power conservation approach if the question only asks for $I_s$ .

Input power = $V_p \times I_p = 220 \times 4 = 880$ W

At 100% efficiency, output power is also 880 W.

I_s = \frac{P_{output}}{V_s}

But we still need $V_s$ for this. So the turns-ratio method is actually more direct here. The power method shines when efficiency is less than 100% — for example, if efficiency is 80%, then output power = $0.8 \times 880 = 704$ W, and you solve from there.

Common Mistake

Flipping the current ratio. Many students write $\frac{I_s}{I_p} = \frac{N_s}{N_p}$ by analogy with the voltage equation — but this is wrong. Current ratio is the inverse of the turns ratio: $\frac{I_s}{I_p} = \frac{N_p}{N_s}$ . This appeared directly as a 2-mark CBSE question in 2024. The logic: step-up means higher voltage and lower current. If both voltage and current increased, we’d be getting free energy.

In CBSE 12 boards, transformer questions almost always give you three of the four quantities ( $V_p$ , $V_s$ , $N_p$ , $N_s$ , $I_p$ , $I_s$ ) and ask for the fourth. Set up the ratio equation first, then substitute. If efficiency is given as less than 100%, use $\eta = \frac{V_s I_s}{V_p I_p}$ — don’t forget to apply it before solving for the unknown current.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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