KCET · Physics · Semiconductors
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 \(110 \mathrm{~V}\)
- C zero
- D \(220 \sqrt{2} \mathrm{~V}\)
Answer & Solution
Correct Answer
(D) \(220 \sqrt{2} \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
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
\(\begin{aligned}
&=V_{0}=V_{\text {rms }} \times \sqrt{2} \\
&\text { Given, } V_{\text {rms }}=220 \mathrm{~V} \\
&\therefore \quad V=220 \sqrt{2} \mathrm{~V}
\end{aligned}\)
\(V=\) Peak value of input AC voltage
\(\begin{aligned}
&=V_{0}=V_{\text {rms }} \times \sqrt{2} \\
&\text { Given, } V_{\text {rms }}=220 \mathrm{~V} \\
&\therefore \quad V=220 \sqrt{2} \mathrm{~V}
\end{aligned}\)
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