Question

What is the equivalent capacitance of the three capacitors in Figure $Ex26.28$?

Short answer

The equivalent capacitance of the three capacitors is$25\mu F$.

Step-by-step solution

Given information

We need to find that the equivalent capacitance of three capacitors.

Capacitance of capacitor

The capacitance of the capacitors with capacitance $20\mu F$and $30\mu F$connected in parallel will be given as

$c_{12}=\frac{1}{{\displaystyle \frac{1}{C_1}}+{\displaystyle \frac{1}{C_2}}}=\frac{1}{{\displaystyle \frac{C_1+C_2}{C_1{C}_2}}}=\frac{C_1{C}_2}{C_1+C_2}$

Putting the values we get

$C_{12}=\frac{2·{10}^{-5}·3·{10}^{-5}}{5·{10}^{-s}}=12\mu F$

Now, the capacitance of the circuit is the capacitance of $C_{12}$in parallel with $C_3=13\mu F$. That will be

$C=C_{12}+C_3=12+13=25\mu F$