Question

In the given circuit resistances are $R_1=6\Omega$, $R_2=4\Omega$ and $R_3=4\Omega$. Inductance is of 2mH. $i_1$, $i_2$ and $i_3$ are the current through $R_1$, $R_2$ and $R_3$ respectively at any time. Switch is closed at t=0.

• A. $I_1(t)=\frac{1}{4}+\frac{3}{4}e^{-3200t}$
• B. $I_1(t)=\frac{5}{4}-\frac{1}{4}e^{-3200t}$
• C. $I_2(t)=\frac{5}{4}+\frac{3}{4}e^{-3200t}$

Answer 1

Answer: (B)

B

Answer 2

To find the current $I_1(t)$ after the switch is closed at $t=0$, we follow the standard procedure for analyzing RL transient circuits.

1. Initial Conditions ($t=0^+$):

At $t=0^+$, an inductor acts as an open circuit to sudden changes in current. Since the switch is closed at $t=0$ and there was no current flowing before, the initial current through the inductor $L$ (which is $i_3$) must be zero:

$i_3(0^+) = 0$.

At this instant, the path through $L$ and $R_3$ is effectively an open circuit. Therefore, the current $i_1$ flows through $R_1$ and $R_2$ only.

The equivalent resistance at $t=0^+$ is $R_{eq}(0^+) = R_1 + R_2 = 6\Omega + 4\Omega = 10\Omega$.

The current $I_1(0^+)$ is given by Ohm's Law: $I_1(0^+) = V / R_{eq}(0^+) = 10V / 10\Omega = 1A$.

2. Final Conditions ($t=\infty$, Steady State):

At steady state, the inductor acts as a short circuit.

The circuit becomes $R_1$ in series with the parallel combination of $R_2$ and $R_3$.

The equivalent resistance of $R_2$ and $R_3$ in parallel is:

$R_{23} = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{4\Omega \times 4\Omega}{4\Omega + 4\Omega} = \frac{16}{8}\Omega = 2\Omega$.

The total equivalent resistance at $t=\infty$ is $R_{eq}(\infty) = R_1 + R_{23} = 6\Omega + 2\Omega = 8\Omega$.

The current $I_1(\infty)$ is given by Ohm's Law: $I_1(\infty) = V / R_{eq}(\infty) = 10V / 8\Omega = 5/4 A = 1.25A$.

3. Time Constant ($\tau$):

The time constant for an RL circuit is $\tau = L/R_{th}$, where $R_{th}$ is the Thevenin equivalent resistance seen by the inductor.

To find $R_{th}$, we turn off the independent voltage source (replace it with a short circuit) and look into the terminals where the inductor is connected.

If the 10V source is shorted, $R_1$ and $R_2$ are in parallel. This parallel combination is then in series with $R_3$ (from the perspective of the inductor).

$R_{1||2} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{6\Omega \times 4\Omega}{6\Omega + 4\Omega} = \frac{24}{10}\Omega = 2.4\Omega$.

$R_{th} = R_{1||2} + R_3 = 2.4\Omega + 4\Omega = 6.4\Omega$.

Now, calculate the time constant:

$\tau = \frac{L}{R_{th}} = \frac{2 \text{ mH}}{6.4 \Omega} = \frac{2 \times 10^{-3} \text{ H}}{6.4 \Omega} = \frac{0.002}{6.4} \text{ s} = \frac{1}{3200} \text{ s}$.

So, the exponential term will be $e^{-t/\tau} = e^{-3200t}$.

4. General Expression for Inductor Current $i_3(t)$:

The current through the inductor follows the general form: $i_L(t) = i_L(\infty) + (i_L(0^+) - i_L(\infty))e^{-t/\tau}$.

First, calculate $i_3(\infty)$. At steady state, $I_1(\infty) = 5/4 A$. This current splits between $R_2$ and $R_3$.

Using the current divider rule:

$i_3(\infty) = I_1(\infty) \times \frac{R_2}{R_2 + R_3} = \frac{5}{4}A \times \frac{4\Omega}{4\Omega + 4\Omega} = \frac{5}{4}A \times \frac{4}{8} = \frac{5}{4}A \times \frac{1}{2} = \frac{5}{8}A$.

Now, substitute the values into the general form for $i_3(t)$:

$i_3(t) = \frac{5}{8} + (0 - \frac{5}{8})e^{-3200t} = \frac{5}{8}(1 - e^{-3200t})$.

5. Expression for $I_1(t)$:

We need to find $I_1(t)$. From Kirchhoff's Voltage Law (KVL) in the main loop containing the source, $R_1$, and $R_2$:

$V = I_1 R_1 + I_2 R_2$.

From Kirchhoff's Current Law (KCL) at the node connecting $R_1$, $R_2$, and $L$:

$I_1 = I_2 + I_3 \implies I_2 = I_1 - I_3$.

Substitute $I_2$ into the KVL equation:

$V = I_1 R_1 + (I_1 - I_3)R_2$.

$V = I_1 (R_1 + R_2) - I_3 R_2$.

$10 = I_1 (6\Omega + 4\Omega) - I_3 (4\Omega)$.

$10 = 10 I_1 - 4 I_3$.

Solving for $I_1$:

$10 I_1 = 10 + 4 I_3$.

$I_1 = 1 + 0.4 I_3$.

Now, substitute the expression for $i_3(t)$:

$I_1(t) = 1 + 0.4 \left( \frac{5}{8}(1 - e^{-3200t}) \right)$.

$I_1(t) = 1 + \frac{4}{10} \times \frac{5}{8}(1 - e^{-3200t})$.

$I_1(t) = 1 + \frac{2}{5} \times \frac{5}{8}(1 - e^{-3200t})$.

$I_1(t) = 1 + \frac{10}{40}(1 - e^{-3200t})$.

$I_1(t) = 1 + \frac{1}{4}(1 - e^{-3200t})$.

$I_1(t) = 1 + \frac{1}{4} - \frac{1}{4}e^{-3200t}$.

$I_1(t) = \frac{5}{4} - \frac{1}{4}e^{-3200t}$.

This matches option B.