Question

Two resistors of $10\Omega$ and one $20\Omega$ resistor and an ideal inductor of $10 H$ are connected to a $2 V$ battery as shown in figure. Key $K$ is closed at time $t=0$. The initial $(t=0)$ and final $(t\rightarrow\infty)$ current through the battery are respectively:

• A. $\dfrac{1}{{15}}A,\,\,\,\dfrac{1}{{10}}A$
• B. $\dfrac{1}{{10}}A,\,\,\,\dfrac{1}{{15}}A$
• C. $\dfrac{2}{{15}}A,\,\,\,\dfrac{1}{{10}}A$
• D. $\dfrac{1}{{15}}A,\,\,\,\dfrac{2}{{25}}A$

Answer 1

Initial current = $\frac{2}{25}A$, Final current = $\frac{1}{10}A$

Answer 2

To determine the initial ($t=0$) and final ($t \rightarrow \infty$) currents through the battery, we need to analyze the behavior of the ideal inductor in the circuit at these two time instances.

Circuit Analysis:

From the given figure, the circuit consists of:

1. A 2V battery.
2. A 20Ω resistor (let's call it $R_3$).
3. A parallel combination of three components:
   • A 10H ideal inductor (L).
   • A 10Ω resistor (let's call it $R_1$).
   • Another 10Ω resistor (let's call it $R_2$).

The 20Ω resistor ($R_3$) is in series with this parallel combination.

1. Initial Current (at $t=0$):

At the instant the key $K$ is closed ($t=0$), an ideal inductor behaves as an open circuit because it opposes any sudden change in current. This means no current can flow through the inductor branch at $t=0$.

• Equivalent Circuit at $t=0$:

  Since the inductor branch is open, the current flows only through the two 10Ω resistors ($R_1$ and $R_2$) which are in parallel.

  The equivalent resistance of $R_1$ and $R_2$ in parallel is:

  $R_{p,0} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{10 \, \Omega \times 10 \, \Omega}{10 \, \Omega + 10 \, \Omega} = \frac{100}{20} \, \Omega = 5 \, \Omega$.

  This parallel combination ($R_{p,0}$) is in series with the 20Ω resistor ($R_3$).

  The total equivalent resistance of the circuit at $t=0$ is:

  $R_{eq,0} = R_3 + R_{p,0} = 20 \, \Omega + 5 \, \Omega = 25 \, \Omega$.

• Initial Current Calculation:

  Using Ohm's Law, the initial current through the battery ($I_{initial}$) is:

  $I_{initial} = \frac{V}{R_{eq,0}} = \frac{2 \, V}{25 \, \Omega} = \frac{2}{25} \, A$.

2. Final Current (at $t \rightarrow \infty$):

As time approaches infinity ($t \rightarrow \infty$), the circuit reaches a steady state. In a DC steady state, an ideal inductor behaves as a short circuit (zero resistance) because the current through it becomes constant (i.e., $\frac{dI}{dt} = 0$, so the induced EMF $L\frac{dI}{dt} = 0$).

• Equivalent Circuit at $t \rightarrow \infty$:

  In the parallel combination of the 10H inductor, the 10Ω resistor ($R_1$), and the 10Ω resistor ($R_2$), the inductor now acts as a short circuit. When a short circuit is in parallel with other resistors, the equivalent resistance of that entire parallel combination becomes zero, as all current will flow through the path of zero resistance.

  So, the equivalent resistance of the parallel part is:

  $R_{p,\infty} = 0 \, \Omega$.

  This parallel combination (now effectively a short circuit) is in series with the 20Ω resistor ($R_3$).

  The total equivalent resistance of the circuit at $t \rightarrow \infty$ is:

  $R_{eq,\infty} = R_3 + R_{p,\infty} = 20 \, \Omega + 0 \, \Omega = 20 \, \Omega$.

• Final Current Calculation:

  Using Ohm's Law, the final current through the battery ($I_{final}$) is:

  $I_{final} = \frac{V}{R_{eq,\infty}} = \frac{2 \, V}{20 \, \Omega} = \frac{1}{10} \, A$.

Conclusion:

The initial current through the battery is $\frac{2}{25} \, A$, and the final current through the battery is $\frac{1}{10} \, A$.

The initial $(t=0)$ and final $(t\rightarrow\infty)$ current through the battery are respectively: $\frac{2}{25} \, A$ and $\frac{1}{10} \, A$.

Explanation of the solution:

At $t=0$, the inductor acts as an open circuit. The two 10Ω resistors are in parallel, giving 5Ω. This 5Ω is in series with the 20Ω resistor, making total resistance 25Ω. Current = 2V/25Ω = 2/25 A.

At $t\rightarrow\infty$, the inductor acts as a short circuit. The parallel combination of inductor and two 10Ω resistors becomes a short circuit (0Ω). This 0Ω is in series with the 20Ω resistor, making total resistance 20Ω. Current = 2V/20Ω = 1/10 A.