Question

An electrical heater and a capacitor are joined in series across a $220V,$ $50Hz$ $AC$ supply. The potential difference across the heater is $90V$. The potential difference across the capacitor will be about

• A. $200V$
• B. $130V$
• C. $110V$
• D. $90V$

Answer 1

Answer: (A)

$200V$

The supply voltage $E$, the potential difference $\left(V_R\right)$ across the resistance, and the potential difference $\left(V_C\right)$ across the capacitor are related as

$E^2=V_R^2+V_C^2$
$\therefore V_c=\sqrt{{220}^2-(90{)}^2}={10}^2\sqrt{4.03}$