Question

Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then $T_1$ is connected to $T_2$. As the current in R = 6 $\Omega$ attains a maximum value of steady state level, $T_1$ is disconnected from $T_2$ and immediately connected to $T_3$. Potential drop across r = 3 $\Omega$ resistor immediately after $T_1$ is connected to $T_3$ is _________ V. (Round off to the Nearest Integer) 

 

Hint:

The most important principle for RL circuits with switches is that the current through an inductor cannot change instantaneously. The current just after a switch is thrown is the same as the current just before it was thrown.

Answer 1

Correct Answer: 3

Step 1: Charging Phase ($T_1$ connected to $T_2$).
The inductor L and resistor R are connected to the 6V battery. The current in the circuit increases until it reaches a steady state.
In the steady state, the inductor offers zero resistance to DC current and acts like a simple connecting wire.
The maximum steady state current ($I_{max}$) flowing through the circuit is determined by the resistor R and the battery voltage V.
$I_{max} = \frac{V}{R} = \frac{6 \text{ V}}{6 \, \Omega} = 1$ A.
This is the current flowing through the inductor just before the switch is changed.
Step 2: Decaying Phase ($T_1$ connected to $T_3$).
The switch is moved, disconnecting the battery and connecting the inductor L to the resistor r.
A key property of an inductor is that the current through it cannot change instantaneously.
Therefore, immediately after the switch is connected to $T_3$, the current flowing from the inductor through the resistor r is equal to the steady-state current from the first phase.
Initial current in the decay circuit, $I_0 = I_{max} = 1$ A.
The question asks for the potential drop across the resistor r = 3 $\Omega$ at this exact moment.
Using Ohm's law:
$V_r = I_0 \times r$
$V_r = 1 \text{ A} \times 3 \, \Omega = 3$ V.
The potential drop is 3 V.