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`0.2 F` capacitor is charged to `600 V` by a battery. On removing the battery it is connected with another parallel plate condenser of `1 F`. The potential decreases toA. 100 VB. 120 VC. 300 VD. 600 V |
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Answer» Correct Answer - A total charge q=cv q=0.2 x 600= 120 C total C=C1 +C2= 0.2+1 =1.2 F q=C V 120=1.2 x V V=100 V |
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