Chapter 25 - Capacitance - Problems - Page 740: 11

$3.157\mu F$

Work Step by Step

In the given circuit $C_1$and $C_2$ are in parallel so their equivalent capacitance is $C_{12} = C_1+C_2 = 10\mu F + 5\mu F=15\mu F$ Next, we know that $C_{12}$ and $C_{3}$ are in series so the total equivalent capacitance is; $\frac{1}{C_{equ}}=\frac{1}{C_{12}}+\frac{1}{C_3}$ $\frac{1}{C_{equ}}=\frac{1}{15}+\frac{1}{5}$ $C_{equ} = \frac{15\times 5}{15+5}= 3.157\mu F$

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