Three capacitors, each of capacitance $0.3 ext{ \mu F}$, are connected in parallel. This combination is conn

Question

Three capacitors, each of capacitance $0.3 	ext{ \mu F}$, are connected in parallel. This combination is connected with another capacitor of capacitance $0.1 	ext{ \mu F}$ in series. Then the equivalent capacitance of the combination is:

Accepted Answer

1. **Parallel Combination:** First, calculate the equivalent capacitance ($C_p$) of the three capacitors connected in parallel. For parallel capacitors, capacitances add directly.
   $C_p = C_1 + C_2 + C_3 = 0.3 	ext{ \mu F} + 0.3 	ext{ \mu F} + 0.3 	ext{ \mu F} = 0.9 	ext{ \mu F}$.
2. **Series Combination:** Now, this parallel combination ($C_p = 0.9 	ext{ \mu F}$) is connected in series with the fourth capacitor ($C_4 = 0.1 	ext{ \mu F}$). For series capacitors, the reciprocal of the equivalent capacitance ($C_{eq}$) is the sum of the reciprocals.
   $\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C_4}$
   $\frac{1}{C_{eq}} = \frac{1}{0.9} + \frac{1}{0.1} = \frac{10}{9} + 10 = \frac{10 + 90}{9} = \frac{100}{9}$.
3. **Final Calculation:**
   $C_{eq} = \frac{9}{100} 	ext{ \mu F} = 0.09 	ext{ \mu F}$.

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