Question

In a given network capacitor $C_1=10\mu F,C_2=\mu F,$ , and $C_3=4\mu F$ . The equivalent capacitance between $PQ$ will be

• A. $1.6\mu F$
• B. $3.2\mu F$
• C. $6.4\mu F$
• D. $4.5\mu F$

Answer 1

Answer: (B)

Capacitance $C_1$ and $C_2$ are connected in parallel. Hence, the equivalent capacitance $C_{eq}=C_1+C_2=5+10=15\mu F$ As $C_{eq}$ and $C_3$ are connected in series Hence, effective capacitance between $P$ and $Q$ is given by $\frac{1}{C_{eff}}=\frac{1}{C_{eq}}+\frac{1}{C_3}=\frac{1}{15}+\frac{1}{4}=\frac{19}{60}$ $C_{eff}=\frac{60}{19}\approx 3.2\mu F$