In steady state, a capacitor of capacitance 2 | vedclass

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Question

(C) In steady state, the capacitor acts as an open circuit, so no current flows through the branch containing the capacitor.Therefore, the current $I$

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(C) In steady state, the capacitor acts as an open circuit, so no current flows through the branch containing the capacitor.Therefore, the current $I$ flows only through the $2 \Omega$ resistor.The voltage across the $2 \Omega$ resistor is equal to the terminal voltage $V$ of the cell.Since the capacitor is in parallel with the $2 \Omega$ resistor, the potential difference across the capacitor is equal to the potential difference across the $2 \Omega$ resistor.$V_C = \frac{Q}{C} = \frac{4 	imes 10^{-6} 	ext{ C}}{2 	imes 10^{-6} 	ext{ F}} = 2 	ext{ V}$.Thus, the terminal voltage $V = 2 	ext{ V}$.The current flowing through the $2 \Omega$ resistor is $I = \frac{V}{R} = \frac{2 	ext{ V}}{2 \Omega} = 1 	ext{ A}$.Using the relation for the terminal voltage of a cell, $V = E - Ir$, where $E$ is the emf and $r$ is the internal resistance:$E = V + Ir = 2 	ext{ V} + (1 	ext{ A} 	imes 0.5 \Omega) = 2 + 0.5 = 2.5 	ext{ V}$.Therefore, the emf of the cell is $2.5 	ext{ V}$.

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

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