FAQ
TL;DR: For a 2 V source with two 1 Ω resistors and a 1 A current source between them, "Vab = 0.5 V; Vbc = 1.5 V." Solve using KCL (Ibc = Iab + 1) and KVL (2 = Vab + Vbc). [Elektroda, Steve Lawson, post #21670081]
Why it matters: This FAQ helps students and hobbyists quickly calculate individual voltage drops in mixed source series circuits without sign errors.
Quick Facts
- Solved values here: Vab = 0.5 V, Vbc = 1.5 V; Iab = 0.5 A, Ibc = 1.5 A. [Elektroda, Steve Lawson, post #21670081]
- Correct node relation: Ibc = Iab + 1 A (not Iab = Ibc + 1 A). [Elektroda, Steve Lawson, post #21670081]
- Norton tip: Convert the series voltage source + resistor to a parallel current source + resistor; parallel legs share the same voltage. [Elektroda, Sajed Rakhshani, post #21670080]
- Law check that catches mistakes: 2 − Iab·1 − Ibc·1 = 0 and KCL balance at the node. [Elektroda, Steve Lawson, post #21670076]
- Ohm’s law refresher: V = I·R; use it after solving currents to get drops. [Elektroda, Cody Gass, #21670085
How do I find the voltage drop across each of the two 1 Ω resistors in this mixed-source loop?
Write KCL: Ibc = Iab + 1. Write KVL: 2 = Iab·1 + Ibc·1. Solving gives Iab = 0.5 A and Ibc = 1.5 A. Then Vab = 0.5 V and Vbc = 1.5 V. [Elektroda, Steve Lawson, post #21670081]
What equations should I write first (KCL/KVL) to avoid sign mistakes?
Use KCL at the middle node: Ibc = Iab + 1. Then apply KVL around the loop: 2 = Vab + Vbc = Iab·1 + Ibc·1. Solve for currents, then voltages. “KCL plus KVL nails it.” [Elektroda, Steve Lawson, post #21670081]
Why did someone get 1.5 V across both resistors, and why is that wrong?
That came from a sign error in KCL. If both drops were 1.5 V, KVL would give 3 V, contradicting the 2 V source. The KCL sum at the node would also fail to balance. [Elektroda, Steve Lawson, post #21670076]
What are the actual currents through each resistor here?
From KCL and KVL, Iab = 0.5 A through the left 1 Ω resistor, and Ibc = 1.5 A through the right 1 Ω resistor. That’s a 3× current difference due to the 1 A current source. [Elektroda, Steve Lawson, post #21670081]
How can Norton’s theorem simplify this problem?
Convert the series voltage source and its resistor to a Norton current source in parallel with a resistor. In that equivalent, the two resistors are in parallel and share the same node voltage, clarifying the 1.5 V on the right branch. [Elektroda, Sajed Rakhshani, post #21670080]
What is a Norton equivalent circuit?
It’s an equivalent representation of any linear two-terminal network as a current source in parallel with a resistor. Here, converting helps visualize parallel elements sharing the same voltage. [Elektroda, Sajed Rakhshani, post #21670083]
Can you give me a quick 3-step method to solve similar circuits?
- Write KCL at the critical node(s) to relate branch currents.
- Write KVL around the loop(s) to relate voltages and currents.
- Use Ohm’s law to compute individual drops and verify with KCL/KVL. [Elektroda, Steve Lawson, post #21670073]
How do I compute power in each resistor for this case?
Use the solved values: Vab = 0.5 V and Vbc = 1.5 V across 1 Ω each. Power is Pab ≈ 0.25 W and Pbc ≈ 2.25 W. Higher power on the right branch reflects the larger 1.5 A current. [Elektroda, Steve Lawson, post #21670081]
What common sign convention pitfall should I watch for?
Be consistent with current directions when writing KCL. Swapping the relation to Iab = Ibc + 1 leads to incorrect voltages and KVL failure. “Mind the arrows; math follows.” [Elektroda, Steve Lawson, post #21670076]
If the resistor values change, how do I adapt the solution?
Keep the same equations: KCL at the node and KVL around the loop. Replace 1 Ω with your actual Ra and Rb, solve currents, then use V = I·R to get each drop. [Elektroda, Steve Lawson, post #21670073]
How do I verify my hand calculation quickly?
After solving, check KVL (sum of drops equals the source) and KCL at the node. You can also run a quick simulation; the thread author validated results that way. [Elektroda, Sajed Rakhshani, post #21670080]
What is Ohm’s law and how is it used here?
Ohm’s law states V = I·R. After finding currents from KCL/KVL, multiply by each resistor value to get voltage drops across the resistors. [Elektroda, Cody Gass, post #21670085]
