Consider the circuit shown in the figure. The resistance of the wire used to make the inductor L is negligible compared to the resistors in the circuit. V=16 V, R1=14Ω, R2=105 Ω, and L=2 H.A) Assume that switch S has been closed for a long time so that steady currents exist in the circuit. Find the current being delivered by the battery, the current going through resistor R2 and the current going through the inductor.B) Now let's open the switch. Find the initial voltage across the inductor immediately after we open the switch.C) Let's suppose it has been 0.020 seconds since we opened the switch. Find the current through the inductor. ### Series-Parallel RL Circuit DiagramThe diagram represents a basic electrical circuit containing the following components:1. **Voltage Source (V):** This is represented by the symbol for a battery, providing electrical energy to the circuit.2. **Switch (S):** The switch is in series with the voltage source and the resistor R1, allowing the circuit to be opened or closed. When closed, current flows through the circuit.3. **Resistor (R1):** This is connected in series with the switch. The resistance opposes the flow of current, causing a potential drop.4. **Resistor (R2):** This is connected in parallel to an inductor (L). Similar to R1, R2 opposes current flow but is in a parallel arrangement.5. **Inductor (L):** Connected in parallel with R2, the inductor stores energy in a magnetic field when current flows through it and opposes changes in current.### How the Circuit Works- When the switch (S) is closed, current flows from the voltage source (V) through the resistor R1.- The circuit then splits, allowing current to flow through both the resistor R2 and the inductor L. The parallel configuration causes a shared voltage across R2 and L.- The inductor L creates a magnetic field to store energy, and it resists changes in the flow of current, characteristic of an RL circuit.This basic RL circuit is fundamental in understanding how inductors and resistors interact in circuit design, often used for filtering and timing applications.

Given data is, Here, V = 16 V, R1 = 14 Ω, R2 = 105 Ω , L = 2 H A) In the given circuit, the inductor will behave as a short. Hence, equivalent resistance in the circuit is Req = R1 Req = 14 Ω (a) Total current delivered by the battery is, IB = VReq IB = 1614IB = 1.14 A (b) Here, the inductor behaves as a short circuit. ∴ Resistance R2 is short. Hence the current through the resistance R2 is Zero. (c) Current in the inductor is equal to the current in the resistance R1 IL = VR1 IL = 1614IL = 1.14 AB) When the switch is open, current through the inductor will be same and resistance R1 will have no current. Therefore, the initial voltage across the inductor is, V = ILR2V = (1.14)(105)V = 119.7 Volts C) Value of the time constant is, τ = LRLτ = 2105τ = 0.019 ≈ 0.02 Therefore, current in the inductor after a time of 0.02 sec is, IL(t) = ILe-τR2LIL(t) = (1.27)e-(0.02 × 105)2IL(t) = (1.27)e-1.05IL(t) = 0.444 A