In the capacitor network shown, C₍AB₎/C₍AC₎ = k/m in lowest terms. Find k + m. The six capacitors have values 3 μF, 1 μF, 5 μF, (6/5) μF, 2 μF and 6 μF, connected as shown. [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

In the capacitor network shown, C₍AB₎/C₍AC₎ = k/m in lowest terms. Find k + m. The six capacitors have values 3 μF, 1 μF, 5 μF, (6/5) μF, 2 μF and 6 μF, connected as shown. [See Image.]

G1 Physics · Accepted Answer

Step 1 — Treat each capacitor as a conductance for the node-voltage method; set voltage across A and the selected terminal equal to 1 V. Step 2 — For A–B, floating nodes C and D each settle at 1/3 V. Total charge at A corresponds to C₍AB₎ = 58/15 μF. Step 3 — For A–C, the floating nodes B and D each settle at 1/6 V. Then C₍AC₎ = 29/6 μF. Step 4 — C₍AB₎/C₍AC₎ = (58/15)/(29/6) = 4/5, giving k + m = 4 + 5 = 9.