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A capacitor of capacitance $5 \mu F$ is connected as shown in the figure. The internal resistance of the cell is
$0.5 \Omega$. The amount of charge on the capacitor plate is
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$0.5 \Omega$. The amount of charge on the capacitor plate is
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Verified Answer
The correct answer is:
$10 \mu C$
Potential difference a cross line $(2)=$ potential difference a cross capacitor
$=1 \times 2=2 \mathrm{Vth}$
So, charge on capacitor $=5 \times 2=10 \mu \mathrm{C}$
In steady state condition. No current flows through line (1). Hence total current
$i=\frac{2.5}{(1+1+0.5)}=1 \mathrm{~A}$
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