A) 4 volt
B) 6 volt
C) 8 volt
D) 10 volt
Correct Answer: A
Solution :
\[\frac{1}{{{C}_{eq}}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\Rightarrow {{C}_{eq}}=1\,\mu \,F\] Total charge Q = Ceq.V = 1 × 24 = 24 mC So p.d. across 6 mF capacitor = \[\frac{24}{6}=4\,volt\]You need to login to perform this action.
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