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

Question

Calculate the steady state current in the 2Ω resistor, shown in figure. The internal resistance of the battery is negligible and the capacitance of the capacitor C is 0.2μ F . (A)  1.0 A  (B)  0.9 A  (C)  1.2 A  (D)  0.8 A

Tardigrade Admin · Accepted Answer

As in steady state capacitor acts as open circuit, on current will flow through

4Ω

resistance.

So,

R_{e q} = \frac{2 \times 3}{( 2 + 3 )} + 2.8 = 4 Ω

and hence,

I = \frac{V}{R _{e q}} = \frac{6}{4} = \frac{3}{2} A

Now, is in parallel current divides in the inverse ratio of resistances

l_{2} = \frac{R _{3}}{( R _{2} + R _{3} )} l = \frac{3}{( 2 + 3 )} \times \frac{3}{2}

= 0.9 A

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

$1.0 A$

$0.9 A$

$1.2 A$

$0.8 A$

Q. Calculate the steady state current in the 2Ω resistor, shown in figure. The internal resistance of the battery is negligible and the capacitance of the capacitor C is 0.2μF .

1.0A

0.9A

1.2A

0.8A

As in steady state capacitor acts as open circuit, on current will flow through 4Ω resistance. So, Req​=(2+3)2×3​+2.8=4Ω and hence, I=Req​V​=46​=23​A Now, is in parallel current divides in the inverse ratio of resistances l2​=(R2​+R3​)R3​​l=(2+3)3​×23​ =0.9A

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

$1.0 A$

$0.9 A$

$1.2 A$

$0.8 A$