Question

In figure shown, value of current in $10\Omega$ resistor just after plug of key $K$ is inserted is:

Answer 1

0.1A

Answer 2

To find the current in the 10Ω resistor just after the key K is inserted, we need to analyze the circuit at $t=0^+$ (the instant immediately after the switch is closed).

1. Inductor Behavior at $t=0^+$:
Before the key K is inserted, the circuit is open, meaning no current flows through any component. Therefore, the current through the 1H inductor at $t=0^-$ (just before closing the key) is $I_L(0^-) = 0$.

A key property of an inductor is that it opposes any instantaneous change in current. Thus, the current through the inductor cannot change instantaneously. So, the current through the inductor at $t=0^+$ (just after closing the key) must be the same as at $t=0^-$.
$I_L(0^+) = I_L(0^-) = 0$.
When the current through an inductor is zero, it behaves as an open circuit.

2. Redrawing the Circuit at $t=0^+$:
Since the 1H inductor acts as an open circuit at $t=0^+$, the branch containing the inductor effectively becomes an open path, meaning no current flows through it.

The circuit then simplifies to the 3V battery, the 10Ω resistor, the key K (now closed), and the parallel combination of the 30Ω and 60Ω resistors.

3. Calculate Equivalent Resistance:
The 30Ω and 60Ω resistors are connected in parallel. Their equivalent resistance ($R_p$) is: $R_p = \frac{30 \Omega \times 60 \Omega}{30 \Omega + 60 \Omega} = \frac{1800}{90} \Omega = 20 \Omega$ Now, the 10Ω resistor is in series with this equivalent resistance $R_p$. The total resistance ($R_{total}$) in the circuit, as seen by the battery, is: $R_{total} = 10 \Omega + R_p = 10 \Omega + 20 \Omega = 30 \Omega$

4. Calculate the Current in the 10Ω Resistor:
The total current flowing from the 3V battery will flow through the 10Ω resistor, as the inductor branch is open. Using Ohm's Law: $I = \frac{V}{R_{total}} = \frac{3V}{30 \Omega} = 0.1 A$ Therefore, the current in the 10Ω resistor just after the key K is inserted is 0.1A.