Using nodal analysis, find: (a) the voltage V12; (b) the voltage V10; (c) the vo

Question

Using nodal analysis, find:

(a) the voltage V12;

(b) the voltage V10;

(c) the voltage V20;

(d) the current through R1 flowing in the direction from the node 1 to the ground;

(e) the current through R2 flowing in the direction from the node 2 to the ground;

Explanation / Answer

Define node V_a as the node to the left of the middle 12k resistor and node V_b as the node to the right of the middle 12k resistor Assume current to the right of node V_a leaves KCL @ node V_a: (leaving currents = positive and entering currents = negative) I_x + [ V_a - V_b ] / (12k) = 63mA V_a * [ 1 / (12k) ] + V_b * [ -1 / (12k) ] + I_x = 63mA --------> equation ( 1 ) KCL @ node V_b: (currents leaving = positive and currents entering = negative) V_b / (12k) + I_x = [ V_a - V_b ] / (12k) V_a * [ -1 / (12k) ] + V_b * [ 1 / (12k) + 1 / (12k) ] + I_x = 0 -------> equation ( 2 ) Now find an expression for the dependent source: From node V_a, we know I_x: I_x = V_a / (12k) ------> equation ( 3 ) Substitute equation ( 3 ) into equations ( 1 ) and ( 2 ): for equation ( 1 ) we get: V_a * [ 1 / (12k) ] + V_b * [ -1 / (12k) ] + V_a / (12k) = 63mA V_a * [ 1 / (12k) + 1 / (12k) ] + V_b * [ -1 / (12k) ] = 63mA ----------> equation ( 4 ) for equation ( 2 ) we get: V_a * [ -1 / (12k) ] + V_b * [ 1 / (12k) + 1 / (12k) ] + V_a / (12k) = 0 V_a * [ -1 / (12k) + 1 / (12k) ] + V_b * [ 1 / (12k) + 1 / (12k) ] = 0 ---------> equation ( 5 ) Solving equations ( 4 ) and equation ( 5 ) using any method yields: V_a = 378 V and: V_b = 0 V Now I_o can be found: I_o = V_b / (12k) = 0 / (12k) = 0 A

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