In steady state, calculate energy stored in capacitors shown in Fig. and the rate at which battery supplies energy.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. Since, in steady state no current flows through capacitors, therefore, current through 1 Ω resistor becomes zero.
Current through resistors and charges on capacitors will be as shown in Fig. Applying KVL on mesh ABCHJMGA,
3I + 3I + 2I – 10 + 2I = 0
∴ I = 1 A
Mesh GFELMG, + 
+ (1 × 0) – 10 + 2 I = 0
q 1 = 16 µC
Mesh EDHJLE, + 
+ 2 I + (1 × 0) = 0 q 2 = – 4 µC
Mesh BCHJKB, 3I+2I – 
=0 q 3 = 5 µC
Energy stored in capacitors, U = Σ 
= 
+ 
+ 
= 80.5 × 10 –6 J
Rate of supply of energy by battery is P = EI
= 10 × 1 watt = 10 W
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