Question

Problemuǐ(30 points) Consider the circuit in Figure 1 1Ω 4A Figure I. DC resistive circuit Submit these Numerical Values in the Answer Sheet 5 points:IA 5 points: IA 5 points:VV 5 points: V.V 5 points : Power of the 2 Ω resistor Li 5 points: Power of the current source [W 1.a : Use KCL, KVL, and Ohm's Law to calculate 1 [AJ. I2 IA]. V1 IV] and V2 [V] 工 20-120 A 112 2. L.b: Verify your results by checking that power is conserved. Use your calculated voltages and/or currents to verify that the sum of power of each device (using passive sign convention) is equal to zero.

Answer 1

Given circuit: - applying KCL 1a) I_1 + 4 = I_2 rightarrow 1 also I_1 = 20 - v_2/1 I_1 = 20 - v_2 rightarrow 2 I_2 = v_2/2 using ohm's law substitute 2 & 3 in eqn 1 20 - v_2 + 4 = v_2/2 24 - v_2 = 0.5 v_2 1.5 v_2 = 24 v_2 = 16 rad/s \r\n I_1 = 20 - 16 (from eqn 2) I_1 = 4 A I_2 = v_2/2 = 16/2 I_2 = 8A v_1 = I_1(1) v_1 = 4(1) v_1 = 4 volts v_2 = 16 volts P_2 omega = (I_2 omega)^2(2) = 8^2(2) P_2 omega = 128 W P_4 A = (v_2)(4) = 16(4) P_4 A = 64 W \r\n Power delivered by 4A source is P_4 A = 64 W Power delivered by 20 v source is P_20 v = (20)I_1 = (20)(4) P_200 = 0 W Power absorbed by 2 omega resistor is P_2 omega = I_2^2(2) = 8^2(2) P_2 omega = 128 W Power