Capacitors In Series And Parallel

Question

Connecting two capacitors of 4 microfarad and 12 microfarad one after another in a series arrangement gives a combined value, so what is the equivalent capacitance?

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For capacitors joined in series the reciprocal of the equivalent capacitance equals the sum of the reciprocals, $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$, because the same charge sits on each while the voltages add, as shown in NCERT Class 12, Chapter 2 (Electrostatic Potential and Capacitance). Substituting $C_1 = 4\ \mu F$ and $C_2 = 12\ \mu F$ gives $\frac{1}{C_{eq}} = \frac{1}{4} + \frac{1}{12} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3}$, so $C_{eq} = 3\ \mu F$. The value 16 \mu F is wrong because it simply adds the capacitances as if they were in parallel. The value 8 \mu F is wrong because it takes the arithmetic average of the two values. The value 48 \mu F is wrong because it multiplies them without dividing by the sum. A useful shortcut for just two capacitors in series is the product-over-sum rule, $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{4	imes12}{16} = 3\ \mu F$, which reproduces the same answer. A consistency check confirms the result: the series equivalent must be smaller than the smallest individual capacitor, and $3\ \mu F$ is indeed less than $4\ \mu F$, exactly as expected for a series combination.

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