Question

Calculate the steady state current in the $ 2\Omega $ resistor, shown in figure. The internal resistance of the battery is negligible and the capacitance of the capacitor $ C $ is $ 0.2\mu F $ .

• A. $ 1.0\,A $
• B. $ 0.9\,A $
• C. $ 1.2\,A $
• D. $ 0.8\,A $

Answer 1

Answer: (B)

As in steady state capacitor acts as open circuit, on current will flow through $4 \Omega$ resistance.

So, $ R_{ eq }=\frac{2 \times 3}{(2+3)}+2.8=4 \,\Omega$
and hence, $I=\frac{V}{R_{ eq }}=\frac{6}{4}=\frac{3}{2} \,A$
Now, is in parallel current divides in the inverse ratio of resistances
$l_{2}=\frac{R_{3}}{\left(R_{2}+R_{3}\right)} l=\frac{3}{(2+3)} \times \frac{3}{2}$
$=0.9\, A$