    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 ΩGiven, 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=100−25=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, (7.5−2.5=5 A). Applying Ohm's law across R, we get: 25=5×R ⇒R=5 Ω Hence, option (d) is the correct answer.  Suggest Corrections  0      Similar questions  Explore more