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Q. The steady state current through the battery in the circuit given below isPhysics Question Image

J & K CETJ & K CET 2010Current Electricity

Solution:

In steady state, the branch containing capacitor will become ineffective (ie, we can ignore $1 \Omega$ resistance)
The equivalent circuit will now be
$\therefore R_{eq}=\frac{R_{1} \times R_{2}}{R_{1}+R_{2}}$
$=\frac{5 \times 2}{5 +2}=\frac{10}{7}$
Now, $V=i\, R\, 10=\frac{i \times 10}{7}$
$\Rightarrow i=7\, A$

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