Find the voltage and the current through L1,L2,C1,R1,R2,R3. The image depicts an electrical circuit diagram containing various components that form a complex circuit.### Components:1. **Voltage Sources:** - A 5V DC voltage source. - A 12V DC voltage source.2. **Inductors:** - \( L_1 = 10 \, \text{mH} \) (millihenries). - \( L_2 = 100 \, \text{mH} \) (millihenries).3. **Resistors:** - \( R_1 = 10 \, \text{k}\Omega \) (kilo-ohms). - \( R_2 = 4.7 \, \text{k}\Omega \) (kilo-ohms). - \( R_3 = 10 \, \text{k}\Omega \) (kilo-ohms).4. **Capacitor:** - \( C_1 = 0.1 \, \mu\text{F} \) (microfarads).### Connections:- The 5V source is connected in series with \( L_1 \).- \( L_1 \) is followed by \( R_1 \) and \( R_2 \) in parallel.- The parallel circuit extends to \( R_3 \), connecting one side to ground.- After \( R_2 \), the circuit connects to \( L_2 \) on the right.- \( L_2 \) is in series with the 12V source.- A capacitor \( C_1 = 0.1 \, \mu\text{F} \) is connected in parallel between the ground and the junction of \( R_1 \) and \( R_2 \).This circuit involves parallel and series connections of various components, showcasing a combination of reactive (inductors and the capacitor) and resistive elements, powered by dual DC sources.

Given: Brief description: In the above given question they have mentioned an electrical circuit having independent voltage source, current source, resistor and inductor. From the given question it is very clear that here we have to perform only DC analysis. While considering DC analysis we have to calculate the reactance of the inductor and capacitor. For DC analysis inductor is short circuit when capacitor is open circuited. By the help of KVL and ohm's law we have calculated all the required current and voltages. The given circuit is completely a DC circuit. For DC circuits inductor and capacitor are replaced by the reactances. In DC analysis inductor is short circuited and capacitor is open circuited. For DC analysis all the inductors are short circuited and the voltage across both inductor will be zero volt. VL1 = 0 Volt VL2 = 0 Volt. Let I4 be the current flowing in resistance R3. I4 = V / 10 K I4 = 12 / 10 K I4 = 1.2 m A By KCL IL1 = I4 + IL2 IL2 = IL1 - I4 IL2 = -1.4893 m A - 1.2 m A IL2 = - 2.6893 m A. By KVL 5 - 10 K × 0 - VC1 = 0 VC1 = 5 - 0 VC1 = 5 Volt Since the capacitor is open circuited for DC , hence the value of current and resistance R1 is 0 m A IR1 = 0 m A . Since capacitor is open circuited for the busy analysis hence it will allow zero current to pass through it. IC = 0 m A From the above given circuit the value of current flowing in the resistance 4.7 kilo ohm is same as current flowing through the inductor. That's IL1 = IR2 IL1 = -1.4893 m A IL1 = IR2 IR2 = -1.4893 m A. Answers : IL1 = -1.4893 m A IR1 = 0 m A IC1 = 0 m A IR2 = -1.4893 m A IR3 = I4 = 1.2 m A IL2 = - 2.6893 m A IC = 0 m A. VL1 = 0 Volt. VL2 = 0 Volt