In the circuit shown. The potential drop across each capacitor is (assuming the two diodes are ideal). .A. 12 V,12 VB. 16 V,8 VC. zero, 24 VD. 8 V, zero

In the circuit shown. The potential drop across each capacitor is (ass

1.	In the circuit shown. The potential drop across each capacitor is (assuming the two diodes are ideal). .A. 12 V,12 VB. 16 V,8 VC. zero, 24 VD. 8 V, zero
Answer» Correct Answer - B The diode connected parallel to the battery is reverse bias. So current will not pass through it. So total emf divided among `C_(1)` and `C_(2)` and in the inverse ratio of their capacities `V=(q)/(C),(V_(1))/(V_(2)) =(C_(2))/(C_(1))` `V_(1)=(VC_(2))/(C_(1)+C_(2)),V_(2)=(VC_(1))/(C_(1)+C_(2))`.

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In the circuit shown. The potential drop across each capacitor is (assuming the two diodes are ideal). .A. 12 V,12 VB. 16 V,8 VC. zero, 24 VD. 8 V, zero

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