Capacitors In Series And Parallel

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, $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{4\times 12}{16}=3\mu F$, 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.