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Question

A heater is designed to operate with a power of 1000 W in a 100 V line. It is connected, in combination with a resistance R, to a 100 V main supply as shown in Figure. What should be the value of R such the heater may operate with a power of 62.5 W?


A
50 Ω
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B
25 Ω
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C
15 Ω
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D
5 Ω
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Solution

The correct option is D 5 Ω
Given,

Power of the heater, P=1000 W,

Voltage across it, V=100 V,

So, the resistance of the heater is,

R=V2P=100×1001000=10 Ω


The power on which it operates is P=62.5 W.

So, potential drop across heater,

V=R×P=10×62.5=625=25 V

The potential drop across AB,

VAB=10025=75 V

The current in AB,

I=VABR=7510=7.5 A

This current divides into two parts.
Let I1 be the current that passes through the heater. Therefore,

25=I1×10

I1=2.5 A

So, the current through R is,

(7.52.5=5 A).

Applying Ohm's law across R, we get:

25=5×R

R=5 Ω

Hence, option (d) is the correct answer.

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