Question

A 120 volt A.C line provides power to 48 Watt bulb (rated at 120v). If a capacitor of capacitance $2.5\mu F$ is connected in series with the bulb, then the output power of the bulb is $[\omega ={10}^{3}rad/s]$

• A. 20 Watt
• B. 17.28 Watt
• C. 13.3 Watt
• D. 35.7 Watt

Answer 1

The correct option is B 17.28 Watt

By $p=i^{2}R=vi=\frac{v^2}{R}\Rightarrow R=\frac{v^2}{p}=\frac{{120}^2}{48}=300\Omega$
$x_c=\frac{1}{wc}=\frac{{10}^6}{{10}^3\times 2.5}=400\Omega$
$i=\frac{\epsilon }{z}=\frac{120}{\sqrt{{400}^2+{300}^2}}=\frac{120}{500}$
$P_{out}=i^{2}R=(\frac{120}{500}{)}^2\times 300=17.28Watt$