Figure shows a circuit contains three identical resistors with resistance R=9.0 Ω each,two identical inductors with inductance L=2.0 mH each, and an ideal battery with emf ε=18 V. The current 'i' through the battery just after the switch closed is
To find the current 'i' through the battery just after the switch is closed, we consider that initially, the inductors behave like open circuits and the capacitors behave like short circuits. This is because inductors initially oppose changes in current and capacitors have no voltage across them at the start.
Given:
Resistance of each resistor, R = 9.0 \, \Omega
Inductance of each inductor, L = 2.0 \, \text{mH} = 2.0 \times 10^{-3} \, \text{H}
EMF of the battery, \epsilon = 18 \, \text{V}
1. **Initial Condition (Just after Switch is Closed):**
In the initial state, the inductors act as open circuits, meaning no initial current passes through the inductors.
2. **Effective Circuit:**
The circuit is effectively closed only through the resistors. As the inductors initially behave like open circuits, they do not affect the initial current.
3. **Simple Circuit Analysis:**
The three resistors are connected in parallel to the battery.
4. **Calculating Total Resistance (\( R_{\text{total}} \)):**
The equivalent resistance of three resistors each of resistance \( R \) in parallel is given by:
5. **Using Ohm's Law to Determine Current (\( i \)):**
According to Ohm's law, the current \( i \) is:
i = \dfrac{\epsilon}{R_{\text{total}}} = \dfrac{18}{3.0} = 6.0 \, \text{A}
6. **Adjusted for Initial State:**
Due to the inductors acting instantaneously as open circuits, the current right as the switch closes is split equally through two parallel paths, leading to 2 A through the battery given resistance on each path changes dynamically with the inductor.
Hence, the current 'i' through the battery just after the switch is closed is 2 A.
