Physics · Electrostatic Potential And Capacitance · NEET
Two capacitors are in SERIES if the only path from one to the next passes through a single common junction with nothing else joined there — the same charge Q sits on both. They are in PARALLEL if both their left plates connect to one common point and both their right plates connect to another common point — the same voltage V is across both. Trick: follow the wire. If a node connects only two capacitor plates and nothing else, those two are in series.
In series the same charge Q sits on every capacitor, and the total voltage is the sum: V = Q/C1 + Q/C2, so 1/C = 1/C1 + 1/C2. In parallel the same voltage V is across each, and the total charge adds: Q = C1·V + C2·V, so C = C1 + C2. This is exactly opposite to resistors, which is why students mix it up.
No, that is correct. In series the equivalent capacitance is always LESS than the smallest capacitor in the chain (e.g. 3 uF and 6 uF in series give 2 uF). In parallel the equivalent is always MORE than the largest one. Use this as a quick sanity check on your answer.
Check if the bridge is balanced: it is balanced when C1/C2 = C3/C4 for the four arm capacitors. If balanced, the two ends of the middle (bridge) capacitor are at the same potential, so no charge sits on it — simply remove it and solve the remaining four as two series branches in parallel. NEET 2024 used exactly this trick.
Redraw it. Label every node (junction) with a letter. Capacitors sharing the same pair of node labels are in parallel; combine them first. Then look for a lone node joining only two capacitors — those are in series. Replace each reduced block with a single equivalent capacitor and repeat until one capacitor remains between the two terminals.
The equivalent capacitance of the system shown in the circuit (a 3 uF and a 6 uF capacitor connected end to end in series across the terminals) is:
In a circuit, four 2 uF capacitors form the four arms of a bridge between terminals A and B, with a fifth 2 uF capacitor as the bridge (middle) branch. The equivalent capacitance between A and B is:
Five capacitors C1 = C2 = C3 = C4 = 10 uF and C5 = 2.5 uF are connected in a symmetric network across a 50 V battery. The equivalent capacitance and the charge on each capacitor respectively are:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For two capacitors in series, C = (C1 x C2)/(C1 + C2) — the product over the sum. For example, 3 uF and 6 uF give (3x6)/(3+6) = 18/9 = 2 uF. Note the answer is smaller than the smaller capacitor.
For capacitors in parallel you simply add them: C = C1 + C2. For example, 3 uF and 6 uF in parallel give 9 uF. The parallel answer is always larger than the largest capacitor.
Reduce it in steps. Combine any clearly parallel groups first (add them), then combine series chains (reciprocal sum), redraw the simpler circuit, and repeat until one capacitor is left between the terminals. Labeling the junctions with letters makes it easy to spot which capacitors share the same two nodes.
A four-arm bridge is balanced when C1/C2 = C3/C4 for the arm capacitors. When balanced, the bridge (middle) capacitor has equal potential on both ends, carries zero charge, and can be removed. Then the four arms reduce as two series branches in parallel.
In series all capacitors carry the SAME charge Q, but different voltages. In parallel all capacitors have the SAME voltage V, but different charges. This is the key idea behind the series and parallel formulas.
