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Question

Three capacitors C1, C2 and C3 of capacitance 4 μF, 3 μF and 6 μF are charged upto a potential difference of 5 V, 6 V and 4 V respectively. If terminal a is connected with f, terminal e is connected with d and b is conncted with c, then find the amount of heat produced in the circuit?


A
20 μJ
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B
2 μJ
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C
152 μJ
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D
150 μJ
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Solution

The correct option is D 150 μJ
Let initial charges on caapcitors C1, C2 and C3 are q1, q2 and q3 respectively.
q1=4×5=20 μC,
q2=6×3=18 μC,
q3=6×4=24 μC.


Let q amount charge flows in the circuit,


Applying KVL, to the loop PQRS,

24q6+18q3+20q4=0

482q+724q+6003q=0

180=9q

q=20 μC

Now the charges on capcitors C1, C2 and C3 respectively,

q1=2020=0

q2=|1820|=2 μC

q3=2420=4 μC

We know that,
Heat produced (H)=UiUf

Where, initial stored energy, Ui=U1+U2+U3

Ui=q212C1+q222C2+q232C3

Ui=2022×4+1822×3+2422×6

Ui=50+54+48=152 μJ

Final stored energy,
Uf=U1+U2+U3

Uf=022×4+222×3+422×6

Uf=2 μJ

H=UiUf=1522=150 μJ

Hence, option (d) is correct option.

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