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Question

(D) The circuit consists of three $1 \, \mu F$ capacitors in series forming an upper branch,which is in parallel with a single $1 \, \mu F$ capacitor

Vedclass · Accepted Answer

$1.33$

Vedclass · Answer

(D) The circuit consists of three $1 \, \mu F$ capacitors in series forming an upper branch,which is in parallel with a single $1 \, \mu F$ capacitor connected directly between $P$ and $Q$.First,calculate the equivalent capacitance of the three capacitors in series $(C_s)$:$\frac{1}{C_s} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 3 \, \mu F^{-1} \implies C_s = \frac{1}{3} \, \mu F$.Now,this $C_s$ is in parallel with the fourth $1 \, \mu F$ capacitor $(C_p)$:$C_{PQ} = C_s + C_p = \frac{1}{3} \, \mu F + 1 \, \mu F = \frac{4}{3} \, \mu F \approx 1.33 \, \mu F$.Thus,the correct option is $D$.

Four capacitors are connected as shown in the figure. The equivalent capacitance between the points $P$ and $Q$ is ....... $\mu F$.

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