The circuit shown here has two batteries of $8.0\,V$ and $16.0\,V$ and three resistors $3\,\Omega ,\,9\,\Omega $ and $9\,\Omega $ and a capacitor of $5.0\,\mu F.$
How much is the current $I$ in the circuit in steady state? ................... $A$
JEE MAIN 2014, Diffcult
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In steady state capacitor is fully charged hence no current will flow through line $2$ .
Bysimplyfing the circuit
Hence resultant potential difference across resistances will be $8.0 \,\mathrm{V}$
Thus current $1=\frac{V}{R}=\frac{8.0}{3+9}=\frac{8}{12}$
or, $\quad I=\frac{2}{3}=0.67\, \mathrm{A}$
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