A) 22\[\mu \]C
B) 100\[\mu \]C
C) 2800\[\mu \]C
D) 200\[\mu \]C
Correct Answer: D
Solution :
As the capacitors are connected in series. Hence, their equivalent capacitance is given by \[\frac{1}{{{C}_{eq}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}=\frac{1}{3}+\frac{1}{10}+\frac{1}{15}\] \[=\frac{10+3+2}{30}\] \[\frac{1}{{{C}_{eq}}}=\frac{15}{30}=\frac{1}{2}\Rightarrow {{C}_{eq}}=2\mu F\] As the charge on each capacitor will be same \[={{C}_{eq}}V=2\times 100=20\mu C\]
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