In a circuit with a dependent current source whose value is 2 I_x in parallel with a 5 Ω resistor, where I_x is the current through a 2 Ω resistor connected to the same node, how would you express I_x and set up the nodal equation?

In a circuit with a dependent current source whose value is 2 I_x in parallel with a 5 Ω resistor, where I_x is the current through a 2 Ω resistor connected to the same node, how would you express I_x and set up the nodal equation?

• A. I_x = V/2; The dependent source supplies a current of 2 I_x = V; KCL at the node: V/5 + V = I_injected
• B. I_x = V/2; The dependent source supplies 2 I_x = V/2; KCL: V/5 + V/2 = I_injected
• C. I_x = 2V; The dependent source supplies 4V; KCL: V/5 + 4V = I_injected
• D. I_x = V; The dependent source supplies 2 I_x = 2V; KCL: V/5 + 2V = I_injected

Answer: (A)

The main idea is to relate everything to the node voltage V using Ohm’s law and then apply Kirchhoff’s current law at the node. First, the current through the 2 Ω resistor is I_x = V/2, since it’s connected from the node to ground. The dependent current source has value 2 I_x, which becomes 2*(V/2) = V. If we take the current through this source as leaving the node (i.e., delivering current away from the node toward the rest of the circuit), its contribution to the current leaving the node is V. The 5 Ω resistor carries a current leaving the node of V/5. By KCL, the sum of currents leaving the node equals the current entering from the external injection, I_injected. Therefore the nodal equation is V/5 + V = I_injected. This matches the given choice because it uses I_x = V/2, the dependent source current 2 I_x = V, and the KCL relation with the currents leaving the node.

The main idea is to relate everything to the node voltage V using Ohm’s law and then apply Kirchhoff’s current law at the node.

First, the current through the 2 Ω resistor is I_x = V/2, since it’s connected from the node to ground.

The dependent current source has value 2 I_x, which becomes 2*(V/2) = V. If we take the current through this source as leaving the node (i.e., delivering current away from the node toward the rest of the circuit), its contribution to the current leaving the node is V.

The 5 Ω resistor carries a current leaving the node of V/5.

By KCL, the sum of currents leaving the node equals the current entering from the external injection, I_injected. Therefore the nodal equation is V/5 + V = I_injected.

This matches the given choice because it uses I_x = V/2, the dependent source current 2 I_x = V, and the KCL relation with the currents leaving the node.