A transformer having efficiency of $75$ % is working on $220 V$ and $4.4 k W$ power supply. If the current in the secondary coil is $5 A$ . What will be the voltage across secondary coil and the current in primary coil?

V_{s} = 220 V

i_{p} = 20 A

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V_{s} = 660 V

i_{p} = 15 A

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V_{s} = 660 V

i_{p} = 20 A

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V_{s} = 220 V

i_{p} = 15 A

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Solution

The correct option is B $V_{s} = 660 V$ , $i_{p} = 20 A$
Given, $P_{i n} = 4.4 k W$ , $V_{P} = 220 V$ , $i_{S} = 5 A$ , $V_{S} =$ ?
We know that, $i_{P} = \frac{P_{i n}}{V_{P}} = \frac{4.4 \times 10^{3}}{220} = 20 A$
$∴$ Efficiency, $η = \frac{O u t p u t p o w e r}{I n p u t p o w e r}$
Percentage efficiency $η = \frac{P_{o u t}}{P_{i n}} \times 100$
$75$ % $= 100 \times \frac{P_{o u t}}{4.4 \times 10^{3}}$
$∴ P_{o u t} = \frac{75 \times 4.4 \times 10^{3}}{100}$
$= 3300 W$
$∴ P_{o u t} = V_{S} \cdot i_{S}$
$\Rightarrow V_{S} = \frac{3300}{5} = 660 V$

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A transformer having efficiency of 75% is working on 220V and 4.4kW power supply. If the current in the secondary coil is 5A. What will be the voltage across secondary coil and the current in primary coil?

The correct option is B Vs=660V, ip=20AGiven, Pin=4.4kW, VP=220V, iS=5A, VS= ?We know that, iP=PinVP=4.4×103220=20A∴ Efficiency, η=OutputpowerInputpowerPercentage efficiency η=PoutPin×10075% =100×Pout4.4×103∴Pout=75×4.4×103100 =3300W∴Pout=VS⋅iS⇒VS=33005=660V

A transformer having efficiency of $75$ % is working on $220 V$ and $4.4 k W$ power supply. If the current in the secondary coil is $5 A$ . What will be the voltage across secondary coil and the current in primary coil?