A) \[2\mu F\]
B) \[4\mu F\]
C) \[8\mu F\]
D) \[16\mu F\]
Correct Answer: C
Solution :
\[\operatorname{Voltage} across two plates = \frac{{{q}_{net}}}{{{C}_{net}}}\] \[40=\frac{2\times 200}{2\times \operatorname{C}}\] \[2+C=10\] \[C=8\mu \operatorname{F}\]You need to login to perform this action.
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