First slide

Electrical power

Question

The resistance of a heater coil is 110 ohm. A resistance R is connected in parallel with it and the combination is joined in series with a resistance of 11 ohm to a 220 volt main line. The heater operates with a power of I10 watt. The value of R in ohm is

Moderate

A

12.22
B

24.42
C

Negative
D

That the given values are not correct

Solution

Power consumed by heater is 110 W, so by using
$P = \frac{V^{2}}{R}$
$110 = \frac{V^{2}}{110} \Rightarrow V = 110 V .$ Also from figure, $i_{1} = \frac{110}{110} = 1 A$ and $i = \frac{110}{11} = 10 A$ . $So i_{2} = 10 - 1 = 9 A$
Applying Ohm's law for resistance R, V = iR
$\Rightarrow 110 = 9 \times R \Rightarrow = 12 .22 Ω$

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