Question

A 110 V, 60 W lamp is run from a 220 V AC mains using a capacitor in series withe the lamp, instead of a resistor then the voltage across the capacitor is about:

• A. 110 V
• B. 190 V
• C. 220 V
• D. 311 V

Answer 1

The correct option is B 190 V

The correct option is similar to B

Given,

$110V,60W$ lamp is run from a $220V$

The lamp is close to an ideal resistance giving voltage $\left(Vr\right)$ at phase angle $\theta =0^o$;

The capacitor is close to an ideal reactance giving voltage $\left(Vc\right)$

at phase angle $\theta ={90}^o$

The voltages across the lamp and capacitor will be in quadrature with one another,

and the supply voltage $\left(Vs\right)$ will be the PHASOR SUM of $VrandVc$

$\Rightarrow V{s}^2=V{r}^2+V{c}^2$

$\Rightarrow V{c}^2=V{s}^2-V{r}^2$

$\Rightarrow V{c}^2={220}^2-{110}^2$

$\Rightarrow V{c}^2=36300$

$\Rightarrow Vc=\sqrt{36300}$

$\Rightarrow Vc\approx 191V$