**Question 10**Assume the switch is closed and enough time has elapsed for the circuit to be in **steady-state,** and the components are ideal. Find the value of \( V_1 \) for the circuit below. Merely write the number; the answer is understood to be in volts.**Circuit Description:**The circuit includes the following components:1. A 12 V DC voltage source connected to a switch.2. An 8 Ω resistor in series with the voltage source.3. A parallel branch with: - A 2 H inductor and a 4 Ω resistor in series, with a voltage across the resistor labeled \( V_1 \). - A 100 µF capacitor in series with an 8 Ω resistor, with a voltage across the resistor labeled \( V_2 \).**Analysis:**- In a steady-state condition, the inductor acts as a short circuit, and the capacitor acts as an open circuit.- For the left branch (with the inductor and resistor), the current \( I_1 \) flows through the 4 Ω resistor.- In the right branch, there is no current through the capacitor, and only \( I_2 \) flows through the 8 Ω resistor.The task is to find \( V_1 \), the voltage across the 4 Ω resistor in the left branch when the circuit reaches steady state.

In circuit under steady state condition after the switch is closed, the capacitor will act as open circuit and the inductor will act as open circuit. The equivalent circuit in steady state is shown below: From the equivalent circuit, the 12 V voltage source, 8Ω resistor and 4Ω resistor forms a closed loop and left-hand side 8Ω resistor in open circuited.To calculate the voltage across 4Ω resistor, apply voltage division rule in the closed loop: Hence, the answer of the blank is to be filled with “4”.