Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

0.9 \mu F

0.09 \mu F

0.1 \mu F

0.01 \mu F

Step-by-Step Solution

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

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