Figure 30-27 shows a circuit with two identical resistors and an ideal inductor.

Is the current through the central resistor more than, less than, or the same as that through the other resistor (a) just after the closing of switch S, (b) a long time

after that, (c) just after S is reopened a long time later, and (d) a long time after that?

1. Fig.30-27,
2. Both resistors are identical,
3. There is ideal inductance in the circuit.

Using the properties of resistors in parallel and in series, the properties of the inductors and applying to the given Fig.30-27, find the current through the central resistor more than, less than, or the same as that through the other resistor in just after the closing of switch S, a long time after that, just after S is reopened a long time later, and a long time after that.

Formulae are as follow:

$i=\frac{V}{R}$

Where,

V = potential difference,

R = resistance,

i = current.

From Fig.30-27, if switch S is close, just after that according to the Ohm’s law, the current in the central resistance is given by,

$i=\frac{V}{R}$

Also, there is the inductor connected as shown in Fig.30-27 and just after the closing of switch S, the induced emf$\epsilon$ is generated in the inductor. Because of this, the induced emf$\epsilon$, the current in the other resistance is a little bit more than zero.

Therefore, the current in the central resistor is more.

Hence, just after the closing of switch S, the current through the central resistor is more than that through the other resistor.

A long time after closing the switch S, the inductor had reached its steady state and it start to act like a simple conducting wire. Since two given resistances are parallel and both are identical. Thus, the current coming from the battery splits equally between both resistors.

Therefore, the current in both the resistors is the same.

Hence, a long time after closing the switch S, the current through the central resistor is the same as that through the other resistor.

Just after S is reopened along time letter the current in the inductor drops a little bit and induced emf$\epsilon$ on the inductor apposes to the change in the current through it. Now the circuit is equivalent to an imaginary battery with emf$\epsilon$ and the resistors are in series.

Therefore, the current in both resistors is the same.

Hence, just after S is reopened a long time later, the current through the central resistor is the same as that through the other resistor.

A long time after reopening the switch, the current through both resistors is zero.

Therefore, the current in both the resistors is the same.

Hence, a long time after reopening the switch S, the current through the central resistor is the same as that through the other resistor.