question_answer

When a potential difference of \[{{10}^{3}}V\]is applied between A and B, a charge of 0.75 mC is stored in the system of capacitors as shown. The value of C is ( in \[\mu F\])

A)  \[\frac{1}{2}\]     

B)  \[2\]     

C)  \[2.5\]    

D)  \[3\]

Correct Answer: B

Solution :

: Points A and B are not mentioned in the question paper. Assuming the points A and B as follows The equivalent circuit of the given network is as shown in the figures. Hence, the equivalent capacitance between A and B \[{{C}_{eq}}=\frac{1(C+1)}{1+C+1}=\frac{C+1}{C+2}\] The charge stored on the system of capacitors is \[Q=0.75mA=0.75\times {{10}^{-3}}A\] As \[Q={{C}_{eq}}V\] \[\therefore \] \[0.75\times {{10}^{-3}}=\frac{(C+1)}{(C+2)}\times {{10}^{3}}\] On solving, we get \[C=2\]