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Question

The resistance of a heater coil is 110 Ω. A resistance R is connected in parallel with it and combination is joined in series with a resistance of 11 Ω to a 220 V main line. The heater operates with a power of 110 W. The vale of R in ohm is

A
12.22
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B
24.42
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C
negative
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D
that the given value is not correct
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Solution

The correct option is A 12.22
The power across heater coil is
P=V2hRh
V2h=PRh
V2h=110×110
Vh=110V
Since resistance R and coil are in parallel voltage across them is same.
Voltage across 11Ω resistor is also 110 V.
Now, current across 11Ω resistor is
I=11011
I=10A
Also, current through coil is
Ih=VhRh
Ih=110110
Ih=1A
Same current equal to current through 11Ω resistor will flow through the parallel combination of coil and R. Hence, current through R is IIh=101=9A
Now, resistance R is
R=VRIR=1109=12.22Ω

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