A circuit containing a two position switch S is shown in figure.
(a) The switch S is in position 1.

Q: Why is the current in the capacitor branch zero in part (a)?

In a DC circuit, once a capacitor is fully charged, it blocks any further flow of direct current, acting as an open circuit in the steady state.

Q: How do we determine the signs in the KVL equations?

The sign depends on the direction of the loop traversal. A voltage drop across a resistor is negative if traversing in the direction of current, and a battery's voltage is positive if traversing from the negative to the positive terminal.

Q: Why does the inductor branch not affect the circuit in position 1?

When the switch is in position 1, the inductor branch is open at one end, meaning it does not form a closed loop and no current can flow through it.

Q: What happens to the current when the switch is moved to position 2?

The circuit changes to an L-R series circuit. The current doesn't jump to its steady value instantly due to the inductor's opposition to changes in current; instead, it grows exponentially.

Q: How is the time constant calculated for the L-R circuit?

The time constant $\tau$ is the ratio of the equivalent inductance to the equivalent resistance of the active loop, $\tau = L / R_{eq}$.

Question

A circuit containing a two position switch S is shown in figure.
(a) The switch S is in position 1. Find the potential difference $V_A - V_B$ and the rate of production of joule heat in $R_1$.
(b) If now the switch S is put in position 2 at $t = 0$. Find
(i) steady current in $R_4$ and
(ii) the time when current in $R_4$ is half the steady value. Also, calculate the energy stored in the inductor $L$ at that time.

Sigma Percentile · Accepted Answer

This problem is a classic example of a multi-state circuit where the configuration changes dynamically based on a switch. It tests your ability to apply Kirchhoff's laws in steady-state DC conditions and your understanding of transient responses in L-R circuits.

### Part A: Steady State Analysis (Switch at Position 1)

When the switch S is in position 1, the inductor branch is completely disconnected from the active power sources. Furthermore, since we are analyzing the steady state, the capacitor $C$ acts as an open circuit. This means no direct current can flow through the top-right branch containing the capacitor and $R_5$.

We are left with a two-loop circuit. Let's define two clockwise mesh currents: $i_1$ for the bottom loop and $i_2$ for the top-left loop. Applying Kirchhoff's Voltage Law (KVL) to these loops yields:

**Loop 1 (Top-Left):**
$$-2i_2 + 2(i_1 - i_2) + 12 = 0$$

**Loop 2 (Bottom):**
$$-12 - 2(i_1 - i_2) + 3 - 2i_1 = 0$$

Simplifying these equations gives us a system of linear equations:
1. $i_1 - 2i_2 = -6$
2. $4i_1 - 2i_2 = -9$

Solving these simultaneously, we find the mesh currents to be $i_1 = -1	ext{ A}$ and $i_2 = 2.5	ext{ A}$.

To find the potential difference $V_A - V_B$, we trace the path from node A to node B. According to the circuit's polarity and the direction of our mesh currents:
$$V_A + 3 - 2i_1 = V_B$$
$$V_A - V_B = 2i_1 - 3 = 2(-1) - 3 = -5	ext{ V}$$

The rate of Joule heating in resistor $R_1$ depends on the net current flowing through it, which is $i_1 - i_2 = -1 - 2.5 = -3.5	ext{ A}$. The power dissipated is:
$$P_{R1} = (i_1 - i_2)^2 R_1 = (-3.5)^2 	imes 2 = 24.5	ext{ W}$$

### Part B: Transient Analysis (Switch at Position 2)

When the switch is moved to position 2 at $t=0$, the upper part of the circuit is bypassed. The 3V battery ($E_2$) is now connected directly in series with $R_2$, the inductor $L$, and $R_4$. This forms a simple L-R series circuit.

The equivalent resistance of this new loop is $R_{eq} = R_2 + R_4 = 2 + 3 = 5\,\Omega$. The steady-state current $i_0$ that will eventually flow through $R_4$ is simply:
$$i_0 = \frac{E_2}{R_{eq}} = \frac{3}{5} = 0.6	ext{ A}$$

Because of the inductor, the current does not reach this steady value instantly. It grows exponentially according to the equation $i = i_0(1 - e^{-t/	au})$. We need to find the time $t$ when the current is half of its steady value ($i = 0.5 i_0$).

$$0.5 i_0 = i_0(1 - e^{-t/	au}) \implies e^{-t/	au} = 0.5 \implies t = 	au \ln 2$$

The time constant $	au$ for the circuit is $\frac{L}{R_{eq}} = \frac{10 	imes 10^{-3}}{5} = 2	ext{ ms}$. Substituting this back:
$$t = 2 	imes 10^{-3} 	imes 0.693 = 1.386 	imes 10^{-3}	ext{ s}$$

Finally, the energy stored in the magnetic field of the inductor at this exact moment (when $i = 0.3	ext{ A}$) is:
$$U = \frac{1}{2} L i^2 = \frac{1}{2} (10 	imes 10^{-3}) (0.3)^2 = 4.5 	imes 10^{-4}	ext{ J}$$

Visualized Solution

$t \to \infty$ \text{ with Switch at 1}

\text{Kirchhoff's Voltage Law (KVL)}

\text{Solving for Currents}

\text{Potential Difference } V_A - V_B

\text{Joule Heating in } R_1

\text{Switch in Position 2}

\text{Steady Current in } R_4

\text{Current Growth in L-R Circuit}

\text{Energy Stored in Inductor}

The Sigma Insight: L-R Circuit

Part A

Part B