A power transmission line feeds input power at 2.3 kV to a step-d… — Solution | JEE Physics

MCQEasyJEE 2024AC Generator & Transformer

A power transmission line feeds input power at $2.3 \, \text{kV}$ 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 the transformer is made of copper. The output current of the transformer is:

A
$45 \, \text{A}$
B
$50 \, \text{A}$
C
$60 \, \text{A}$
D
$55 \, \text{A}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Input voltage $V_{\text{in}} = 2300 \, \text{V}$ , input current $I_{\text{in}} = 5 \, \text{A}$ , efficiency $\eta = 0.9$ , and output voltage $V_{\text{out}} = 230 \, \text{V}$ .

Find: The output current $I_{\text{out}}$ .

Using input power,

P_{\text{in}} = V_{\text{in}} I_{\text{in}} = 2300 \, \text{V} \times 5 \, \text{A} = 11500 \, \text{W}

Using transformer efficiency,

P_{\text{out}} = \eta P_{\text{in}} = 0.9 \times 11500 \, \text{W} = 10350 \, \text{W}

Now apply output power relation,

P_{\text{out}} = V_{\text{out}} I_{\text{out}}

So,

I_{\text{out}} = \frac{P_{\text{out}}}{V_{\text{out}}} = \frac{10350 \, \text{W}}{230 \, \text{V}} = 45 \, \text{A}

Therefore, the output current of the transformer is $45 \, \text{A}$ . The correct option is A.

Using Efficiency Formula Directly

Given: $V_p = 2300 \, \text{V}$ , $I_p = 5 \, \text{A}$ , $V_s = 230 \, \text{V}$ , and $\eta = 90\% = 0.90$ .

Find: Secondary current $I_s$ .

For a transformer,

\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{V_s I_s}{V_p I_p}

Substitute the given values,

0.90 = \frac{230 \times I_s}{2300 \times 5}

Rearranging,

I_s = \frac{0.90 \times 2300 \times 5}{230}

I_s = 0.90 \times 10 \times 5 = 45 \, \text{A}

The number of primary turns $3000$ is not required for this calculation. Therefore, the output current is $45 \, \text{A}$ , so the correct option is A.

Common mistakes

Using the ideal transformer relation and ignoring efficiency is incorrect here because the transformer is only $90\%$ efficient. First calculate output power as $P_{\text{out}} = \eta P_{\text{in}}$ , then find the secondary current.
Using $2.3$ instead of $2300 \, \text{V}$ causes a unit-conversion error. Convert $2.3 \, \text{kV}$ to volts before computing input power.
Trying to use the given primary turns $3000$ is unnecessary for this question. The required current follows directly from power and efficiency relations, so focus on $P = VI$ .

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