Please solve Q no 16

 

Asked by Nishantthakre95 | 15th Sep, 2019, 05:53: AM

Expert Answer:

Initially capacitors C1 and C2 are charged in the loop ABEF , before switch S is closed.
 
By considering capacitance of C1 and C2 are 2μF and by applying Kirchoff's voltage law for the loop ABEF ,
 
we have,   -10 + [q / (2×10-6)] -10 + [q / (2×10-6)]  = 0   
 
Hence charge on each of capacitors C1 and C2 ,     q = 20 × 10-6 C
 
Hence potential across each capacitance = q/C = 20 × 10-6 / 2× 10-6 = 10 V
 
When switch is closed, capacitor C3 is charged in the loop BGDE.
 
By Kirchoff's voltage in the loop BGDE,  -20 + [q / (2×10-6)] -10 = 0   
 
Hence charge on capacitor C3 ,   q = 60× 10-6 C
 
To charge the capacitors, workdone by battery,  q × V = 60× 10-6 × 20 = 1200 μJ
 
Energy stored in capacitor = (1/2)CV2 = (1/2)×2× 10-6 × 30 × 30 = 900 μJ
 
Heat energy produced in the circuit = (1200 - 900 ) μJ = 300 μJ  = 0.3 mJ
 

Answered by Thiyagarajan K | 15th Sep, 2019, 12:28: PM

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