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Alternating Current question

2018 · 16 Apr · Shift 1 · Q56

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JEE Main
Physics
Alternating Current
2018 · 16 Apr · Shift 1 · Q56

Alternating Current question

2018 · 16 Apr · Shift 1 · Q56

JEE MainPhysicsAlternating CurrentMCQ+4 / −1

A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns, giving the output power at 230 V. If the current in the primary of the transformer is 5 A, and its efficiency is 90%, the output current would be ;

A
50 A
B
45 A
C
25 A
D
20 A

View written solutionFree

Correct answer: B

Given data

Primary voltage: $V_p = 2300\,\text{V}$
Secondary voltage: $V_s = 230\,\text{V}$
Primary current: $I_p = 5\,\text{A}$
Efficiency of transformer: $\eta = 90\% = 0.9$

(Primary turns are given as $4000$ , but they are not needed here since voltages are already given.)

Input power to transformer

$P_{\text{in}} = V_p I_p = 2300 \times 5 = 11500\,\text{W}$

Output power using efficiency

Efficiency is

$\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$

So,

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

Find output current

Using

$P_{\text{out}} = V_s I_s$

$I_s = \frac{P_{\text{out}}}{V_s} = \frac{10350}{230} = 45\,\text{A}$

Check options

A: $50\,\text{A}$ ❌
B: $45\,\text{A}$ ✅
C: $25\,\text{A}$ ❌
D: $20\,\text{A}$ ❌

Therefore, the correct answer is Option B.

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