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Alternating Current question

2019 · 9 Jan · Shift 2 · Q71

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JEE Main
Physics
Alternating Current
2019 · 9 Jan · Shift 2 · Q71

Alternating Current question

2019 · 9 Jan · Shift 2 · Q71

JEE MainPhysicsAlternating CurrentMCQ+4 / −1

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

A
50 A
B
45 A
C
35 A
D
25 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: $\eta = 90\% = 0.9$

We need to find the secondary (output) current $I_s$ .

Input power to transformer

The input electrical power is

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

Output power using efficiency

Efficiency is defined as

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

So,

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

Find output current

Output power is also

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

Therefore,

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

Match with options

A: $50\,\text{A}$
B: $45\,\text{A}$
C: $35\,\text{A}$
D: $25\,\text{A}$

Hence, the correct option is:

$\boxed{\text{B: }45\,\text{A}}$

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.

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