In steady state, a capacitor of capacitance $2 \mu \mathrm{F}$ is charged to $4 \mu \mathrm{C}$, as shown in figure. If the internal resistance of the cell is $0.5 \Omega$, then the emf of the cell is

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Aicharya · Accepted Answer

$ \mathrm{V}_C=\frac{Q}{C}=\frac{4 	imes 10^{-6}}{2 	imes 10^{-6}}=2 \mathrm{v} $ Now, $\mathrm{V}_C=\mathrm{V}$ (across $2 \Omega$ resistor) $ 	herefore \quad I=\frac{V}{R}=\frac{2}{2}=1 \mathrm{~A} $ Now, using the relation, $\mathrm{V}=E-I r$ So, $\mathrm{E}=V+I r=2+1 	imes 0.5=2.5 \mathrm{~V}$

In steady state, a capacitor of capacitance $2 \mu \mathrm{F}$ is charged to $4 \mu \mathrm{C}$, as shown in…

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