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The power dissipated in the circuit shown in the figure is 30 Watts. The value of R is
A
30 $\Omega$
B
20 $\Omega$
C
15 $\Omega$
D
10 $\Omega$
Detailed Solution
R and the 5 $\Omega$ resistor are in parallel across the 10 V battery.
Total power: $\frac{V^2}{5} + \frac{V^2}{R} = 30 \Rightarrow \frac{(10)^2}{5} + \frac{(10)^2}{R} = 30$
$20 + \frac{100}{R} = 30 \Rightarrow \frac{100}{R} = 10$
$R = 10\ \Omega$
Total power: $\frac{V^2}{5} + \frac{V^2}{R} = 30 \Rightarrow \frac{(10)^2}{5} + \frac{(10)^2}{R} = 30$
$20 + \frac{100}{R} = 30 \Rightarrow \frac{100}{R} = 10$
$R = 10\ \Omega$