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Question

In the Fig 20.46 shown the potential drop across 3μF capacitor when switch S is open and switch s is closed is :

144735_58650659bb964d77a2a5ba24018d8880.png

A
9 V, 8 V
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B
9 V, 9 V
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C
6 V, 8 V
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D
12 V, 8 V
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Solution

The correct option is C 12 V, 8 V
When switch S is open, 18V will drop across C1=3μF and C2=6μF.
Thus, potential across 3μF is V1=VC2C1+C2=18×63+6=12V.
When the switch S is closed potential drop across 8Ω resistor is the potential drop across 3μF.
The current through each resistor is I=188+10=1A.
Thus voltage across 8Ω resistor V=IR=1×8=8V
Therefore , the potential drop across 3μF=8V.

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