A 220 V A.C. supply is connected between points A and B (figure). What will be the potential difference V across the capacitor? - Physics

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MCQ

A 220 V A.C. supply is connected between points A and B (figure). What will be the potential difference V across the capacitor?

Options

  • 220 V

  • 110 V

  • O V

  • `220 sqrt(2)` V

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Solution

`220 sqrt(2)` V

Explanation:

Half wave rectifier: When the P-N junction diode rectifies half of the ac wave, it is called a half-wave rectifier.


(i) During positive half cycle,

Diode → forward biased

Output signal → obtained

(ii) During the negative half cycle,

Diode → reverse biased

Output signal → not obtained

(iii) Output voltage is obtained across the load resistance RL. It is not constant but pulsating (mixture of ac and dc) in nature.

(iv) Average output in one cycle

`I_(dc) = I_0/pi` and `V_(dc) = V_0/pi, I_0 = V_0/(r_f + R_L)`  .....(`r_f` = forward based resistance)

(v) r.m.s. output: `I_(rms) = I_0/2, V_(rms) = V_0/2`

As p-n junction diode will conduct during the positive half cycle only, during negative half cycle diode is reverse biased. During this diode will not give any output. So, the potential difference across capacitor C = peak voltage of the given AC voltage = `V_0 = V_(rms) sqrt(2) = 220 sqrt(2)  V`

Concept: Application of Junction Diode as a Rectifier
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Chapter 14: Semiconductor Electronics - Exercises [Page 88]

APPEARS IN

NCERT Exemplar Physics Class 12
Chapter 14 Semiconductor Electronics
Exercises | Q 14.04 | Page 88

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