Lesson 3 Flashcards — Kirchhoff’s Laws

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1. Define node, branch, and loop.

A branch is a two-terminal element. A node is a connection of two or more elements — everything joined by plain wire is the same node. A loop is any closed path.

2. State Kirchhoff's Voltage Law.

The algebraic sum of voltages around any closed loop is zero: \(\sum V_n = 0\). Physically, the energy a charge gains at the source is exactly the energy it gives up going around.

3. State Kirchhoff's Current Law.

The algebraic sum of currents at any node is zero: \(\sum I_n = 0\), or equivalently \(\sum I_\text{in} = \sum I_\text{out}\). Charge cannot pile up at a junction.

4. When writing a KVL equation, does it matter which sign convention you pick?

No — you may use the first sign you encounter or the second, as long as you are consistent for the whole loop. Mixing the two mid-problem is what produces wrong answers.

5. Where does the + go when you label a resistor's voltage polarity?

Where the current enters the resistor. Current flows from high to low potential through a passive element.

6. You guessed a current direction and got a negative answer. What does that mean?

That the current actually flows the other way. The magnitude is still correct — a negative result is information, not an error, so there is no need to redo the problem.

7. How do you check a circuit answer using conservation of power?

Power supplied by the sources must equal power absorbed by the loads: \(\sum P_\text{supplied} = \sum P_\text{absorbed}\). Compute \(P = IV\) for every element and compare. It costs thirty seconds and catches most sign errors.

8. Which law do you reach for when you know currents and want voltages?

KVL. Conversely, KCL when you know voltages and want currents. Most real problems need both, plus Ohm's Law to connect them.

9. How many independent equations do you need to solve a circuit?

As many as there are unknowns. Write KVL around enough loops and KCL at enough nodes to reach that count, then use Ohm's Law to relate voltages and currents.

10. A source drives 4.8 A at 24 V. What power does it supply?

\(P = IV = (4.8)(24) = 115.2\ \text{W}\). If the loads do not also absorb 115.2 W, there is an error to find.