In steady state, a capacitor of capacitance 2 μ F is charged to 4 μ C, as shown in figure. If the internal resistance of the cell is 0.5 Ω, then the emf of the cell is

Question

In steady state, a capacitor of capacitance 2 μ F is charged to 4 μ C, as shown in figure. If the internal resistance of the cell is 0.5 Ω, then the emf of the cell is (A) 4 V (B) 5 V (C) 2.5 V (D) 2 V

Tardigrade Admin · Accepted Answer

V text cap =(Q/C)=(4 × 10-0/2 × 10-6) =2 v Now V text cap = V (across 2 Ω resistor) ∴ I=(V/R)=(2/2)=1 A Now, using the relation E =V+I r =2+1 × 0.5 =2.5 V

Q. In steady state, a capacitor of capacitance $2 μ F$ is charged to $4 μ C$ , as shown in figure. If the internal resistance of the cell is $0.5 Ω$ , then the emf of the cell is

4 V

5 V

2.5 V

2 V

$V_{cap} = \frac{Q}{C} = \frac{4 \times 1 0 ^{- 0}}{2 \times 1 0 ^{- 6}}$
$= 2 v$
Now $V_{cap} = V$ (across $2 Ω$ resistor)
$∴ I = \frac{V}{R} = \frac{2}{2} = 1 A$
Now, using the relation
$E = V + I r$
$= 2 + 1 \times 0.5$
$= 2.5 V$

Q. In steady state, a capacitor of capacitance 2μF is charged to 4μC, as shown in figure. If the internal resistance of the cell is 0.5Ω, then the emf of the cell is

4 V

5 V

2.5 V

2 V

Vcap ​=CQ​=2×10−64×10−0​ =2v Now Vcap ​=V (across 2Ω resistor) ∴I=RV​=22​=1A Now, using the relation E=V+Ir =2+1×0.5 =2.5V

Q. In steady state, a capacitor of capacitance $2 μ F$ is charged to $4 μ C$ , as shown in figure. If the internal resistance of the cell is $0.5 Ω$ , then the emf of the cell is

$V_{cap} = \frac{Q}{C} = \frac{4 \times 1 0 ^{- 0}}{2 \times 1 0 ^{- 6}}$
$= 2 v$
Now $V_{cap} = V$ (across $2 Ω$ resistor)
$∴ I = \frac{V}{R} = \frac{2}{2} = 1 A$
Now, using the relation
$E = V + I r$
$= 2 + 1 \times 0.5$
$= 2.5 V$