Three capacitors each having capacitance \(\mathrm{C}=1 \mu \mathrm{F}\) are connected as shown in figure…

Three capacitors each having capacitance \(\mathrm{C}=1 \mu \mathrm{F}\) are connected as shown in figure with a battery of emf \(6 \mathrm{~V}\). If the switch is closed

1. the amount of the charge flowing through the cell will be \(2 \mu \mathrm{C}\)
2. the heat generated in the circuit is \(6 \mu \mathrm{J}\)
3. the amount of the charge flowing through the cell will be \(3 \mu \mathrm{J}\)
4. the energy absorbed by the battery after closing the switch is \(3 \mu \mathrm{J}\)

Solution

When switch is open


When switch is closed

Total charge \(\mathrm{q}_{1}=\frac{2}{3} \mathrm{CE}\)
\(\mathrm{q}_{2}=\frac{1}{2} \mathrm{CE}\)
Hence charge flowing through the cell
\(\Delta \mathrm{q}=\frac{2}{3} \mathrm{CE}-\frac{1}{2} \mathrm{CE}=\frac{1}{6} \mathrm{CE}\)
\(\Delta \mathrm{q}=\frac{1}{6} \mathrm{cv}=1 \mu \mathrm{C}\), Hence work done on the battery \(\mathrm{W}=\Delta \mathrm{q} \mathrm{E}=6 \mu \mathrm{J}\) and heat generated in the circuit \(=6 \mu \mathrm{J}\)

Asked in: JEE Mains - Capacitance - Chapter Test

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