A capacitor of capacitance 5 μ F is connected as shown in the...

A capacitor of capacitance

5 \mu \mathrm{F}

is connected as shown in the figure. The internal resistance of the cell is

0.5 \Omega

. The amount of charge on the capacitor plates is A capacitor of capacitance 5 μ F is connected as shown in the figure. The internal resistance of the cell is 0.5 . The

A capacitor of capacitance 5 μ F is connected as shown in the figure. The internal resistance of the cell is 0.5 . The

Options

A $80 \mu \mathrm{C}$
B $40 \mu \mathrm{C}$
C $20 \mu \mathrm{C}$
D $10 \mu \mathrm{C}$

Correct answer

D. $10 \mu \mathrm{C}$

Step-by-step solution

In steady state, there will be no current in the capacitor branch. Net resistance of the circuit

R=1+1+0.5=2.5 \Omega

Current drawn from the cell

i=\frac{V}{R}=\frac{2.5}{2.5}=1 \mathrm{~A} Potential drop across two parallel branches

\begin{aligned} V & =E-\text{ ir }=2.5-1 \times 0.5 \\ & =2.5-0.5=2.0 \mathrm{~V}\end{aligned}

So, charge on the capacitor plates

q=C V=5 \times 2=10 \mu

\mathrm{C}

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