Four capacitors each of capacitance 16 μ F are connected as shown in the figure. The capacitance between points A and B is : _________ (in μ F).

Solution

Physics sir · 4 steps

Do not trust the straight line of capacitors: label every junction A or B first, then read the circuit.

1. Find the B points. The top wire joins the left end of the row, the junction between the second and third capacitors, and the right end of the row. Point B hangs off this same wire, so all three of these junctions are at the potential of B.
2. Find the A points. The bottom wire joins the junction between the first and second capacitors and the junction between the third and fourth capacitors. Point A hangs off this wire, so both of these junctions are at the potential of A.
3. Read each capacitor. Reading left to right, the junctions go B, A, B, A, B. So every one of the four capacitors has one plate on A and the other on B. All four are in parallel between A and B.

   C_(AB) = 4C = 4 × 16 μF = 64 μF

4. So. The capacitance between A and B is 64 μF.

The common slip: 4. Taking the four capacitors in a row as a series chain gives 16/4 = 4 μF. The connecting wires make every one of them sit directly across A and B.

The idea to keep. When wires loop back across a row of capacitors, give every junction the name of the terminal it is wired to. Any capacitor with one plate on A and the other on B is in parallel with the rest.

Answer from the NTA final answer key. Solution by PrepWiser.