Question

Sol. Key concept: Reduce a resistor ladder successively from its far end.

Between E and M, the path through F and N contains three 2 Ω resistors in series, giving 6 Ω. This is in parallel with the 3 Ω vertical resistor, so the equivalent between E and M is 2 Ω.

$$R_{EM} = \frac{3 \times 6}{3 + 6} = 2\ \Omega$$

Between D and L, the right-hand path is again 2 + 2 + 2 = 6 Ω, in parallel with 3 Ω, giving 2 Ω. Repeating once more gives R_CK = 2 Ω.

$$R_{AB} = 2 + 2 + 2 = 6\ \Omega,\ \ \ \ \ I = \frac{18}{6} = 3\ A$$

$$P = VI = 18(3) = 54\ W$$

The whole 3 A passes through AC. The potential difference between C and K is 3 × 2 = 6 V, so current in the 3 Ω CK branch is 6/3 = 2 A. Therefore (a), (c) and (d) are true; (b) is false.