Question

A $500\Omega$ resistor and a capacitor $C$ are connected in series across $50Hz$. $AC$ supply mains. The r.m.s potential difference recorded on voltmeter $V_1$ and $V_2$ are $V_1=120V$ and $V_2=160V$. The power taken from the mains is:

• A. $480W$
• B. $240W$
• C. $28.8W$
• D. $14.4W$

Answer 1

Answer: (C)

$28.8W$

Given

$f=50Hz$

$V_1=120V$ rms

$V_2=160V$ rms

We know,

$\cos \varphi =\frac{R}{2}$

and power $=V_{rms}{I}_{rms}\cos \varphi$

$=\frac{V_{rms}}{\overline{Z}}{I}_{rms}R=(I_{rms}{)}^{2}R$

$V_{rms}=\frac{V_1}{R}=\frac{120}{500}=\frac{12}{50}$ (as some rms current flows in series component)

Power $={\left(\frac{12}{50}\right)}^2\times 500$

$=\frac{1440}{50}=28.8$ watts

Option C.