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 \[C=2\]You need to login to perform this action.
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