Three capacitors 2μF 3μF 6μF in series and parallel — find equivalent

Question

Three capacitors of capacitances $C_1 = 2\,\mu\text{F}$ , $C_2 = 3\,\mu\text{F}$ , and $C_3 = 6\,\mu\text{F}$ are connected. Find the equivalent capacitance when they are connected (a) in series, and (b) in parallel.

Solution — Step by Step

For capacitors in series, the reciprocals add:

\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

Why? In series, the same charge $Q$ sits on each capacitor, but each has a different voltage. The total voltage $V = V_1 + V_2 + V_3$ , and since $V = Q/C$ , dividing through by $Q$ gives the reciprocal sum formula.

\frac{1}{C_s} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}

Finding common denominator (LCM of 2, 3, 6 is 6):

\frac{1}{C_s} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1

C_s = 1\,\mu\text{F}

The series equivalent is 1 μF — always smaller than the smallest individual capacitor (2 μF here).

For capacitors in parallel, the capacitances add directly:

C_p = C_1 + C_2 + C_3

Why? In parallel, all capacitors share the same voltage $V$ . Each capacitor holds charge $Q_i = C_i V$ . The total charge $Q = Q_1 + Q_2 + Q_3 = (C_1 + C_2 + C_3)V$ , so the equivalent capacitance $C_p = Q/V = C_1 + C_2 + C_3$ .

C_p = 2 + 3 + 6 = 11\,\mu\text{F}

The parallel equivalent is 11 μF — always larger than the largest individual capacitor (6 μF here).

Why This Works

The parallel rule for capacitors is opposite to that for resistors. Resistors in series add directly; capacitors in parallel add directly. This is because capacitance measures the ability to store charge — more plates in parallel means more total “storage area,” so total capacitance increases.

In series, capacitors share the same charge but divide the voltage — the “weakest link” in terms of capacitance limits the overall storage, so the effective capacitance is less than any individual value.

Quick sanity checks: Series $C_{eq}$ is always less than the smallest capacitor; Parallel $C_{eq}$ is always more than the largest capacitor. Use these as instant verification after calculating.

Alternative Method

For the series case with these three specific values (2, 3, 6), notice that $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$ exactly. This is a “nice” number problem — CBSE and JEE often design questions where the reciprocals add to a whole number. Always check if the numbers were chosen to give a clean answer before going through heavy fraction arithmetic.

Common Mistake

In the series formula, students calculate $\frac{1}{C_s} = 1$ and then write $C_s = 1/1 = 1$ correctly — but in harder problems they forget to take the reciprocal at the end. After finding $\frac{1}{C_s}$ , you must flip it to get $C_s$ . If $\frac{1}{C_s} = \frac{3}{4}$ , then $C_s = \frac{4}{3}\,\mu\text{F}$ , not $\frac{3}{4}\,\mu\text{F}$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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