In the given circuit determine
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[ R = 10 Ω Ω
1A in each
V 3 = 3V, V 2 = 2V, V 4 = 4V
10 W
(e) 1 W
(f) 9W
(g) 9V
(h) 4 Ω Ω resistance
(i) 3 W.]
Sol. R eq = R 1 + R 2 + R 3 + r = 10 Ω Ω
i = 
= 
= 1 A
V 3 Ω Ω = 1 × 3 = 3 V, V 4 Ω Ω = 1 × 4 = 4V, V 2 Ω Ω = 1 × 2 = 2V
P consumed = εi = 10 × 1 = 10 W
(e) P generated =i 2 r = 1 W inside the battery
(f) P output = (ε – ir)i = 10 – 1 = 9 W
(g) V battery = ε – ir = 9 V
(h) In series, P 
R 4 Ω Ω consumes maximum power 4 Ω Ω
(i) P 3 Ω Ω = i 2 R = 3 W.
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