Two identical capacitors $$C1$$ and $$C2$$ of equal capacitance are connected as shown in the circuit. Terminals a and b of the key k are connected to charge capacitor $$C1$$ using battery of emf V volt. Now disconnecting a and b the terminals b and c are connected. Due to this, what will be the percentage lossof energy?

Question

Two identical capacitors $$C1$$ and $$C2$$ of equal capacitance are connected as shown in the circuit. Terminals a and b of the key k are connected to charge capacitor $$C1$$ using battery of emf V volt. Now disconnecting a and b the terminals b and c are connected. Due to this, what will be the percentage lossof energy?

Infinity Learn · Accepted Answer

$$Ui=12CV2On$$ switching key at point $$cq0-qc=qc$$ $$2q=q0$$ $$q=q02$$ $$Uf=12q022$$×$$1C+12q022$$×$$1C$$ $$Uf=q024C$$ $$Uf=14CV2$$ $$loss=Ui-UfUi$$×$$100$$ $$=12-14CV212CV2$$×$$100$$ $$=$$ $$50$$%

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