MCQ
A 220 V A.C. supply is connected between points A and B (Fig.). What will be the potential difference V across the capacitor?
- A220V.
- B110V.
- C0V.
- D$220\sqrt{2}\text{V}.$
✓
Answer
- $220\sqrt{2}\text{V}.$
Solution:
Key concept: Half wave rectifier: When the P-N junction diode rectifies half of the ac wave, it is called half wave rectifier.
- During positive half cycle,
Diode → Forward biased
Output signal → obtained
- During negative half cycle,
Diode → reverse biased
Output signal → not obtained
- Output voltage is obtained across the load resistance RL. It is not constant but pulsating (mixture of ac and dc) in nature.
- Average output in one cycle
$\text{I}_\text{dc}=\frac{\text{I}_0}{\pi}\text{ and }\text{V}_\text{dc}=\frac{\text{V}_0}{\pi};\text{I}_0=\frac{\text{V}_0}{\text{r}_\text{f}+\text{R}_\text{I}}$
(rf = forward biased resistance)
- r.m.s. output: $\text{I}_\text{rms}=\frac{\text{I}_0}{2},\text{V}_\text{rms}=\frac{\text{V}_0}{2}$
As p-n junction diode will consuct during positive half cycle only, during negative half cycle diode is reverse biased. During this diode will not give any output. So, potential difference across capacitor C = peak voltage of the given AC voltage
$=\text{V}_0=\text{V}_\text{rms}\sqrt{2}=220\sqrt{2}\text{V}$
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