Consider the circuit shown, where V1V1 = 1.8 V, V2V2 = 2.40 V, R1R1 = 2.2 kΩ, R2R2 = 2.5 kΩ, and R3R3 = 1.5 kΩ.1)What is the current through resistor R1R1 in milliamperes? 2)What is the current through resistor R2R2 in milliamperes? 3)What is the power dissipated in resistor R3R3 in milliwatts? 4)What is the total power in milliwatts delivered to the circuit by the two batteries? ### Explanation of the Circuit DiagramThe diagram represents a basic electrical circuit containing two voltage sources and three resistors arranged in a combination of series and parallel configurations.#### Components:1. **Voltage Sources:** - **V1**: Positioned on the left side of the circuit, supplying power in the clockwise direction. - **V2**: Located on the right side of the circuit, also supplying power in the clockwise direction.2. **Resistors:** - **R1**: Situated in the middle, connected in parallel to the other resistors. - **R2**: Positioned at the bottom left of the circuit, connected in series with V1 and V2. - **R3**: Positioned at the bottom right, in series with R2 and parallel to R1.#### Configuration:- **Series Connection**: R2 and R3 are connected in series with each other but in parallel with R1, forming two possible paths for current flow.- **Parallel Connection**: The central resistor, R1, is in parallel with the series combination of R2 and R3, creating a typical mixed-series-parallel circuit.In this setup, Kirchhoff's laws (current and voltage laws) can be applied to analyze the current distribution and voltage drops across each component. This circuit model is useful for understanding concepts of electrical resistance and power distribution in mixed circuit configurations.

Given, V1 = 1.8 VV2 = 2.40 VR1 = 2.2 kΩR2 = 2.5 kΩR3 = 1.5 kΩ.Now by using nodal analysis at node (a), according to which summation of the net current passing through node (a) is zero ∑I = 0 hence, VaR1 + Va - V1R2 + Va-V2R3 = 0Va2.2 + Va - 1.82.5 + Va - 2.401.5 = 03.75Va + 3.3Va - 5.94 + 5.5Va - 13.2 = 012.55Va= 19.14Va = 1.525 Va) Current through resistance, R1 I1 = VaR1 = 1.5252.2×103 I1 = 0.693 ×10-3 AI1 = 0.693 mA b) Current through resistance, R2 I2 = Va - V1R2 = 1.525 - 1.82.5×103 I2 = -0.11 ×10-3 AI2 =- 0.11 mA the negative sign indicates the outgoing current c) Current through resistance, R3 I3 = Va - V2R3 = 1.525 - 2.401.5×103 I3 = -0.583 ×10-3 AI3 = -0.583 mA The power dissipated in the resistor, R3 P = I32R3 = 0.583×10-321.5×103P = 0.509×10-3 WP = 0.51 mW d) Net power is delivered to the circuit by the two batteries, P = I2V1 + I3V2P = 0.11×10-3 A1.8 V +0.583×10-3 A2.40 VP = 1.5972 ×10-3 WP = 1.592 mW