A) 3 : 1
B) 5 : 2
C) 2 : 3
D) 1 : 1
Correct Answer: C
Solution :
Key Idea: The capacitors having same charge through them will be in series combination. If a source is connected between points P and R, same charge will How through two capacitors in arms PQ and Q.R. Similarly, same charge will flow through capacitors in arms PT, TS and SK. So, equivalent capacitance of left side \[C=\frac{C\times C}{C+C}=\frac{C}{2}\] and equivalent capacitance of right side \[\frac{1}{C}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}=\frac{3}{C}\] \[\therefore \] \[C=\frac{C}{3}\] Now, C and C will in parallel combination, hence, \[{{C}_{1}}=C+C=\frac{C}{2}+\frac{C}{3}=\frac{5C}{6}\] Similarly, if a source is connected between points P and Q, then equivalent capacitance \[{{C}_{2}}=C+\frac{C}{4}\] \[=\frac{5C}{4}\] Hence, the required ratio is given by \[\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{5C/6}{5C/4}=\frac{2}{3}\]
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