Find the current through the resistor R3 given that V=12.0V, R1=4.73Ω, R2=9.46Ω, R3=23.65Ω, and R4=14.19Ω.Group of answer choices0.5585 A4.2279 A5.5585 A2.1545 A0.2495 A The image depicts an electrical circuit diagram featuring a combination of series and parallel resistors. Explanation of the Circuit Diagram:1. **Voltage Source (V):** On the left side of the diagram, there is a voltage source denoted by 'V'. This provides the electrical energy to the circuit.2. **Series Resistor (R1):** - **R1** is connected directly in series with the voltage source. - The current from the voltage source flows through R1 before reaching the parallel section.3. **Parallel Resistors (R2, R3, R4):** - After R1, the circuit splits into three parallel branches, each containing one resistor: R2, R3, and R4. - These resistors are connected side by side, sharing the same two nodes, so the voltage across each is identical. - Each parallel resistor provides an alternate path for the current to flow.4. **Circuit Layout:** - The circuit starts at the voltage source, passes through the series resistor (R1), and then enters the parallel arrangement of R2, R3, and R4. - After flowing through the parallel resistors, the paths converge and the circuit loops back to the voltage source to complete the circuit.This configuration demonstrates how resistors can be arranged in a circuit to control the current and voltage distribution. The total resistance of the parallel section can be calculated using the formula for parallel resistors, and the overall circuit resistance is the sum of R1 and the equivalent resistance of the parallel group.

Given data The potential difference across the battery is V = 12 V The magnitude of the resistance across the first resistor is R1 = 4.73 ohm. The magnitude of the resistance across the second resistor is R2 = 9.46 ohm. The magnitude of the resistance across the third resistor is R3 = 23.65 ohm. The magnitude of the resistance across the fourth resistor is R4 = 14.19 ohm. The expression for the equivalent resistance is given as: Req=R1+11R2+1R3+1R4 Substitute the value in the above equation. Req=4.73 Ω+119.46 Ω+123.65 Ω+114.19 ΩReq=9.307 Ω The expression for the current through the battery is given as: I=VReq Substitute the value in the above equation. I=12 V9.307 ΩI=1.289 AThe expression for the potential difference across the first resistor is given as: V1=IR1 Substitute the value in the above equation. V1=1.287 A×4.73 ΩV1=6.096 V The expression for the potential difference across the third resistor is given as: V3=V-V1 The expression for the current in the third resistor is given as: I3=V3R3I3=V-V1R3 Substitute the value in the above equation. I3=12 V-6.096 V23.65 ΩI3=0.2495 A Thus, the current in the third resistor is 0.2495 A.