Name: Introduction to Electric Current
Uploaded: 2019-07-19
Duration: 36 min 24 s
Description: In this video, we will learn the definition of electric current, its unit and the flow of the current in an electric circuit due to the movement of electrons.

Question

In the circuit shown in figure, switch $S$ is closed at time $t = 0 .$

(A) What is the current $I_{0}$ leaving the battery at $t = 0$ , immediately after the switch is closed?

(B) The current $I$ “long time” later?

(C) What charge has accumulated on the capacitor after this long time?

(D) If, finally, switch $S$ is opened again, how long will it take after the switch is opened for the capacitor to lose $80 %$ of its charge?

Accepted Answer

(A) At $t = 0$ , capacitor acts as short-circuit. There will be no current in the $40 Ω$ resistor.

Therefore, current in the circuit, $I_{0} = \frac{10}{20} = 0.5 A$ .

(B) After long time, in steady state, the capacitor has a voltage across it, but no current flows through the capacitor.

Thus, capacitor acts as open circuit, so current $80 %$ (Resistors are in series).

(C) Voltage across capacitor is equal to voltage across $40 Ω$ resistors, $V_{40} = I R = \frac{1}{6} \times 40$

Charge stored in the capacitor is $Q = C V_{40} = 0.50 \times \frac{1}{6} \times 40 = \frac{10}{3} μC$

(D) Capacitor will discharge through the $40 Ω$ resistor when $S$ is open.

The variation of charge with time $t$ in $R C$ circuit is $q = q_{0} e^{- \frac{t}{R C}}$

After $t = 0 .$ loss in charge, charge left is $\frac{20}{100} q_{0}$ .

So, $\frac{20}{100} q_{0} = q_{0} e^{- (\frac{t}{40 \times 0.5 \times 10^{- 6}})}$

$\frac{1}{5} = \frac{1}{(\frac{t}{e^{20 \times 10^{- 6}}})}$

$t = 20 \times 10^{- 6} \ln (5)$ $= 3.22 \times 10^{- 5} s$

Important Questions on Electric Current and Circuits

Learn and Prepare for any exam you want !