The figure shows a circuit that contains four identical resistors with resistance $$R = 2.0$$ $$\Omega$$, two identical inductors with inductance $$L = 2.0$$ mH and an ideal battery with E.M.F. $$E = 9$$ V. The current $$i$$ just after the switch $$S$$ is closed will be:

Solution & Explanation

1. Identify Behavior of Inductors at $$t = 0$$

An inductor opposes sudden changes in the electrical current passing through it due to Lenz's law. When the switch $$S$$ is closed at time $$t = 0$$ (initial transient state):

• The initial current passing through each inductor must be exactly zero ($$i_L(0^-) = i_L(0^+) = 0$$).
• Consequently, every inductor in the network behaves as an open circuit (infinite resistance block).

2. Simplify the Circuit Network

Let's evaluate the status of the branches containing inductors immediately after closing the switch:

• The center-right branch contains a series combination of an inductor $$L$$ and a resistor $$R$$. Because the inductor acts as an open circuit, no current flows through this path.
• The far-right branch contains a series combination of a resistor $$R$$ and an inductor $$L$$. Similarly, this inductor acts as an open circuit, meaning this path is also completely deactivated.

Removing these two open-circuited branches leaves only the active conductive paths containing resistors:

• The vertical branch closest to the switch containing a single resistor $$R$$.
• The bottom return horizontal wire segment containing a single resistor $$R$$.

These two remaining active resistors are connected in a simple single-loop series configuration across the ideal battery source.

3. Calculate Equivalent Resistance ($$R_{\text{eq}}$$)

Since the two active resistors are connected end-to-end in series, their individual resistance values add up linearly:

$$R_{\text{eq}} = R + R = 2.0 \,\, \Omega + 2.0 \,\, \Omega = 4.0 \,\, \Omega$$

4. Determine the Initial Current ($$i$$)

Using Ohm's Law, we can calculate the total current $$i$$ delivered by the battery with E.M.F. $$E = 9 \,\, \text{V}$$ into this simplified network at $$t = 0^+$$:

$$i = \frac{E}{R_{\text{eq}}}$$

$$i = \frac{9 \,\, \text{V}}{4.0 \,\, \Omega} = 2.25 \,\, \text{A}$$

Correct Option: C ($$2.25 \,\, \text{A}$$)