question_answer

The capacitors \[{{u}_{1}}=1m/s=0.05kg,{{v}_{2}}=30m/s\]have a capacitance \[\Rightarrow \] each and \[{{C}_{2}}\] has capacitance \[{{m}_{1}}\times 1=0.05\times 30\]. The effective capacitance between a and will be:

A) \[\Rightarrow \]              

B)                       \[{{m}_{1}}=1.5kg\]                       

C) \[{{k}_{1}}\]                                      

D) \[{{y}_{1}}\]

Correct Answer: C

Solution :

Key Idea: No charge flows through capacitor \[{{u}_{1}}=1m/s=0.05kg,{{v}_{2}}=30m/s\]. The given network can be replaced through that shown below. This is a wheatstone bridge arrangement in which \[\Rightarrow \] Hence, bridge is balanced. No charge flows through \[{{m}_{1}}\times 1=0.05\times 30\], hence \[\Rightarrow \] is not considered. Capacitors \[{{m}_{1}}=1.5kg\] and \[{{k}_{1}}\] and in series \[{{y}_{1}}\] \[{{k}_{2}}\]  \[{{y}_{2}}\] Also \[y={{y}_{1}}+{{y}_{2}}\] and \[{{k}_{1}}\] are in series, hence \[{{F}_{1}}=-{{k}_{1}}{{y}_{1}}\] Now, C and C" are in parallel, hence net capacitance between a and d is \[{{F}_{2}}=-{{k}_{2}}{{y}_{2}}\]