Sol.

V A = V B = V C = V D = 0

V E = V F = V G = V H = 10V

V I = 10 + 5 = 15 V

V J = 15 V

V K = 10 + 5 = 15 V

V BI = 15 V, V JG = 5 V, V KD = 15 V

Each battery is supplying the current hence each battery is acting as a source.

Let current through BF, CG, HP is respectively i 1 , i 2 , i 3

i 1 = = 15 amp ↓ ,

i 2 = = 2.5 amp ,

i 3 = = 5 amp ↓

For 10 V Battery, current = i 1 – i 2 + i 3 = 15 – 2.5 + 5 = 17.5 amp

(e) P 1 Ω Ω = = = 225 W

P 2 Ω Ω = = 12.5 W

P 3 Ω Ω = = 75 W

Hence, 1 Ω Ω resistance consumes the maximum power.

(f) P I = E I i 1 = 10 × 17.5 = 175 W

P II = E 2 i 1 = 5 × 15 = 75 W

P III = E 3 i 2 = 15 × 2.5 = 37.5 W

P IV = E 4 i 3 = 5 × 5 = 25 W

hence left most battery consume maximum power.