Question

The capacitance $C_{AB}$ in the given network is:

• A. $7\mu F$
• B. $\frac{50}{7}\mu F$
• C. $7.5\mu F$
• D. $\frac{7}{50}\mu F$

Answer 1

The correct option is A $7\mu F$

From the given circuit, we get $\frac{C_1}{C_2}=\frac{5}{10}=\frac{1}{2}$

And $\frac{C_3}{C_4}=\frac{10}{5}=\frac{2}{1}$

Thus $\frac{C_1}{C_2}$ is reciprocal of $\frac{C_3}{C_4}$

For this particular type of circuit, the equivalent capacitance is given by-

$C_{net}=\frac{2C_1{C}_2+C_3(C_1+C_2)}{C_1+C_2+2C_3}$

$C_{net}=\frac{2\left(5\right)\left(10\right)+5(5+10)}{5+10+2\left(5\right)}=\frac{175}{25}=7\mu F$