Question

A diode is connected to a $220$ V (rms) ac source in series with a capacitor as shown in the diagram. Find the voltage (rms) across the capacitor.

Hint:

The diode charges the capacitor to the peak of the ac input and then blocks discharge. Peak voltage is \(\sqrt{2}\) times the rms value.

Answer 1

Step 1: Understanding the Concept:
The diode and capacitor form a half-wave peak detector. The diode lets current pass only when the input is positive. The capacitor charges through the diode and is not able to discharge back through it.

Step 2: Key Formula or Approach:
The rms value of an ac voltage is related to its peak value by:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} \quad \Rightarrow \quad V_0 = \sqrt{2}\, V_{rms} \]

Step 3: Find the peak voltage:
Here \(V_{rms} = 220\ \text{V}\), so
\[ V_0 = 1.414 \times 220 \approx 311\ \text{V} \]

Step 4: Find the capacitor voltage:
In the positive half cycle the diode is forward biased, so the capacitor charges up to the peak value \(V_0\). In the negative half cycle the diode is reverse biased and blocks, so the capacitor has no path to discharge (no load is connected).
The capacitor therefore stays at the peak value. This is a steady dc voltage, so its rms value is also the same number.

Final Answer:
The voltage across the capacitor is the peak value of the supply. \[ \boxed{V_C = 220\sqrt{2} \approx 311\ \text{V}} \]