A) 25 V
B) 50 V
C) 100 V
D) 150 V
Correct Answer: C
Solution :
Given, \[{{C}_{1}}=6\mu F,{{C}_{2}}=12\mu F,V=150\,Volt\] Total capacity, \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\] \[=\frac{1}{6}+\frac{1}{12}\] \[=\frac{2+1}{12}=\frac{1}{C}=\frac{3}{12}\Rightarrow C=4\mu F\] Potential of\[12\mu F\]capacitor \[V=\frac{q}{C}\] \[V=\frac{4\times 150}{12}\] \[V=50\text{ }volt\]
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