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Chapter 26: Q. 28 (page 738)

What is the equivalent capacitance of the three capacitors in Figure $E x 26.28$ ?

Short Answer

Expert verified

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

Step by step solution

01

Given information

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

02

Capacitance of capacitor

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

$c_{12} = \frac{1}{\frac{1}{C_{1}} + \frac{1}{C_{2}}} = \frac{1}{\frac{C_{1} + C_{2}}{C_{1} C_{2}}} = \frac{C_{1} C_{2}}{C_{1} + C_{2}}$

Putting the values we get

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

Now, the capacitance of the circuit is the capacitance of $C_{12}$ in parallel with $C_{3} = 13 μ F$ . That will be

$C = C_{12} + C_{3} = 12 + 13 = 25 μ F$

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