A power transmission line feeds input power at 2 . 3kV to a step down transformer with its primary winding having 3000turns . The output power is delivered at 230V by the transformer. The current in the primary of the

Correct answer: 0 Given: Input power, P _ i = 2300 × 5watt Output power, P _ 0 = 2300 × 5 × 0 . 9 Now, we can write η = P _ out P _ in = V _ S I _ S V _ P I _ P ⇒ 0 . 9 = 230 I _ S 2300 × 5 ⇒ I _ S = 45A .

JEE Main 202430 JanuaryEvening Shift

Exam: JEE Main
Year: 2024
Date: 30 January
Shift: Evening Shift
Subject: Physics
Chapter: Alternating Current

JEE Main 2024 · 30 January · Evening Shift · Physics · Alternating Current

A power transmission line feeds input power at 2 . 3kV to a step...

A power transmission line feeds input power at

2 . 3kV

to a step down transformer with its primary winding having

3000\text{turns}

. The output power is delivered at

230V

by the transformer. The current in the primary of the transformer is

5A

and its efficiency is

90 %

. The winding of transformer is made of copper. The output current of transformer is ____

A

Correct answer

Step-by-step solution

Given:
Input power, ${P}_{i} = 2300 \times 5\text{watt}$
Output power, ${P}_{0} = 2300 \times 5 \times 0 . 9$
Now, we can write $η = \dfrac{{P}_{out}}{{P}_{in}} = \dfrac{{V}_{S}{I}_{S}}{{V}_{P}{I}_{P}}$
$⇒ 0 . 9 = \dfrac{230{I}_{S}}{2300 \times 5} ⇒ {I}_{S} = 45A .$

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