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Question

Two capacitors A and B with capacitances 15 μF and 5 μF are charged to potential difference of 5 V and 9 V respectively. Consider a wire connected to each cepacitor as shown (with free ends). An uncharged capacitor 5 μF with lead wire falls on the free ends to complete the circuit. Find how much amount of heat produced in the circuit?


A
67.5 μJ
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B
180 μJ
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C
210 μJ
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D
390 μJ
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Solution

The correct option is C 210 μJ
Initial charge on capacitors A and B are,

QA=15×5=75 μC and QB=5×9=45 μC respectively.

Let q amount of charge flows through the circuit.


Applying KVL to the loop PQRS,

75q15+45q5q5=0

75q+1353q3q=0

210=7q

q=30 μC

Final charge on all three cepacitors as follows,

QA=75q=7530=45 μC

QB=45q=4530=15 μC

QC=30 μC

Heat produced (H)=UiUf

Where,
Initial energy stored in the capacitors,

Ui=UA+UB

Ui=QA22CA+QB22CB

Ui=(75)230+(45)210

Ui=187.5+202.5=390 μJ

Final energy stored:

Uf=UA+UB+UC

Uf=(QA1)22CA+(QB1)22CB+(QC1)22CC

Uf=(45)230+(15)210+(30)210

Uf=67.5+22.5+90=180 μJ

So,

H=UiUf
=390180
=210 μJ

Hence, option (c) is correct answer.

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