A circuit has a dependent voltage source equal to 0.5 times the voltage across a 4 Ω resistor in another branch. How would you incorporate this in nodal analysis?

A circuit has a dependent voltage source equal to 0.5 times the voltage across a 4 Ω resistor in another branch. How would you incorporate this in nodal analysis?

• A. Include a constraint V_source = 0.5 * V_4Ω and use a supernode if needed
• B. Replace dependent source with a fixed equivalent voltage
• C. Treat the dependent source as an independent source
• D. Ignore the dependent source in nodal equations

Answer: (A)

When a voltage source connects two non-reference nodes, you handle it by forming a supernode and adding a constraint that relates the two node voltages through the voltage source. Here the dependent source has a value 0.5 times the voltage across a 4 Ω resistor in another branch, so you keep that relationship as a equation linking the two node voltages on either side of the dependent source. Define the two nodes on either side of the dependent voltage source as V1 and V2, with the chosen polarity of the source giving V1 − V2 equal to 0.5 times the voltage across the 4 Ω resistor (let’s call that V4Ω, defined with its own polarity). This becomes the constraint: V1 − V2 = 0.5 · V4Ω. If the resistors or other elements connect to either node, you write KCL for the supernode that encloses V1 and V2, summing currents through all elements connected to either node while omitting the current through the voltage source itself. You also relate V4Ω to the node voltages of the resistor’s terminals (V4Ω is the difference between those two node voltages). Solving the resulting equations yields the node voltages and, from them, any currents of interest. The other options fail because a dependent voltage source is not a fixed voltage, so you don’t replace it with a fixed equivalent, and it isn’t an independent source. Ignoring it would omit a crucial constraint that governs the circuit behavior.

When a voltage source connects two non-reference nodes, you handle it by forming a supernode and adding a constraint that relates the two node voltages through the voltage source. Here the dependent source has a value 0.5 times the voltage across a 4 Ω resistor in another branch, so you keep that relationship as a equation linking the two node voltages on either side of the dependent source.

Define the two nodes on either side of the dependent voltage source as V1 and V2, with the chosen polarity of the source giving V1 − V2 equal to 0.5 times the voltage across the 4 Ω resistor (let’s call that V4Ω, defined with its own polarity). This becomes the constraint: V1 − V2 = 0.5 · V4Ω. If the resistors or other elements connect to either node, you write KCL for the supernode that encloses V1 and V2, summing currents through all elements connected to either node while omitting the current through the voltage source itself. You also relate V4Ω to the node voltages of the resistor’s terminals (V4Ω is the difference between those two node voltages). Solving the resulting equations yields the node voltages and, from them, any currents of interest.

The other options fail because a dependent voltage source is not a fixed voltage, so you don’t replace it with a fixed equivalent, and it isn’t an independent source. Ignoring it would omit a crucial constraint that governs the circuit behavior.