A) \[\frac{8}{3}volt\]
B) 4 volt
C) 6 volt
D) 8 volt
Correct Answer: D
Solution :
In the given figure, capacitors \[3\mu F\] and \[6\mu F\] are connected in parallel, so, their effective capacitance is \[C=3+6=9\mu F\] Again C and \[4.5\mu F\] are connected in series. Hence, the total effective capacitance is given by \[C=\frac{9\times 4.5}{9+4.5}=3\mu F\] The charge through the circuit is given by \[q=CV=3\times 12=36\mu C\] Now potential difference across \[4.5\mu F\] capacitor is \[=\frac{q}{C}=\frac{36}{4.5}=8volt\]
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