Q171 / 180 · KCET 2020 · Physics
A $220 \mathrm{~V}$ AC supply is connected between points $A$ and $B$ as shown in figure, what will be the potential difference $V$ across the capacitor?
- A$220 \mathrm{~V}$
- B$110 \mathrm{~V}$
- Czero
- Correct answerD$220 \sqrt{2} \mathrm{~V}$
Solution
As $p$ - $n$ junction diode conducts electricity during only positive half-cycle (forward biased condition ). So, the potential difference across the capacitor is
$V=$ Peak value of input AC voltage
$\begin{aligned}
&=V_{0}=V_{\text {rms }} \times \sqrt{2} \\
&\text { Given, } V_{\text {rms }}=220 \mathrm{~V} \\
&\therefore \quad V=220 \sqrt{2} \mathrm{~V}
\end{aligned}$
Hence, the correct answer is option: 'D'.