question_answer

Four identical capacitors are connected as shown in diagram. When a battery of 6 V is connected between A and B, the charge stored is found to be 1.5 \[\mu C\]. The value of \[{{C}_{1}}\] is                                          [Kerala PMT 2005]

A)            \[2.5\mu F\]

B)            \[15\,\mu F\]

C)            \[1.5\mu F\]

D)            \[0.1\mu F\]

Correct Answer: D

Solution :

           The capacitance across A and B \[=\frac{{{C}_{1}}}{2}+{{C}_{1}}+{{C}_{1}}=\frac{5}{2}{{C}_{1}}\] As Q = CV, \[1.5\mu C=\frac{5}{2}{{C}_{1}}\times 6\] Þ \[{{C}_{1}}=\frac{1.5}{15}\times {{10}^{-6}}\] \[=0.1\times {{10}^{-6}}F=0.1\mu F.\]