| In the circuit shown in Figure, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is |
 |
A) 1.3 V
B) 2.3 V
C) 0 V
D) 0.5 V
Correct Answer: B
Solution :
| Option [b] is correct. |
| Explanation: In the middle right of the circuit the capacitor behaves like an open circuit for dc 0.2 mA current, so current will flow from A to B only. Let potential across A and B is V, so by |
| Kirchhoff's loop law, |
| \[{{\operatorname{V}}_{AB}}=(5,000\times 0.2\times {{10}^{-3}})+0.3+5,000\times 0.2\times {{10}^{-3}}\] |
| \[{{\operatorname{V}}_{AB}}=1V+0.3V+1V\] |
| \[{{\operatorname{V}}_{AB}}=2.3V\] |
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