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In the circuit shown, the potential drop across each capacitor is (assuming the two diodes are ideal) |
Answer» Solution :The diode `D_(1)` is reverse biased (open circuit), but the diode `D_(2)` is forward biased (short circuit).  `THEREFORE` the potential of the battery divides ACROSS the TWO capacitors in the inverse ratio of their capacities. i.E. `(V_(1))/(V_(2))=(C_(2))/(C_(1))=(8)/(4)=(2)/(1)` `V_(1)=(2)/(3)E=(2)/(3)xx24=16V,V_(2)=(1)/(3)E=(1)/(3)xx24=8V` |
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