A) \[30.6\mu F\]
B) \[20.33\mu F\]
C) \[6.67\mu F\]
D) \[3.33\mu F\]
Correct Answer: C
Solution :
Key Idea: Capacitors in series have same amount of charge pacitors In the given circuit between points A and B, a capacitor \[{{C}_{1}}\]is connected and all other capacitor \[{{C}_{2}},{{C}_{3}},{{C}_{4}}\] are connected in series. Hence, equivalent capacitance in series is \[\frac{1}{C}=\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}+\frac{1}{{{C}_{4}}}\] \[\frac{1}{C}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\] \[\frac{1}{C}=\frac{1}{5}\] \[\Rightarrow \] \[C=\frac{5}{3}\mu F\] This is in parallel with \[{{C}_{1}}\], \[\therefore \] \[C=C+{{C}_{1}}=\frac{5}{3}+5=6.67\,\,\mu F\]You need to login to perform this action.
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