Electrical fundamentals
Solve a DC circuit, one step at a time
Work through a series–parallel circuit, check the answer with power, and try two practice problems.
What you’ll learn
- Identify series and parallel connections.
- Calculate branch currents and check conservation of power.
Before you start: You should be comfortable with Ohm’s law, fractions and the units V, A and Ω.
In this guide 6 sections
Start with a circuit you can reduce without simultaneous equations: a 4 Ω resistor in series with a parallel pair of 6 Ω and 3 Ω resistors, connected to an ideal 18 V DC source. Assume ideal wires and fixed resistance values. Find the source current, each branch current and the power supplied.
1. Identify the connections before choosing a formula
The source current passes through the 4 Ω resistor before dividing between two branches. The 6 Ω and 3 Ω resistors share the same pair of connection points. That makes them parallel. The 4 Ω resistor carries the combined current and is in series with that parallel combination.
Label the junction after the 4 Ω resistor A and the return junction B. Both parallel resistors connect from A to B. This connection test is more reliable than deciding from the way a circuit is drawn.
2. Reduce the parallel pair
Between A and B
1 / RAB = 1 / 6 + 1 / 3 = 1 / 2
RAB = 2 Ω
Whole circuit: Rtotal = 4 + 2 = 6 Ω.
The parallel result is smaller than either branch resistance. If you got 9 Ω, you added the branch resistances as if they were in series. Redraw the connections before continuing.
3. Find current, then return to the original branches
Ohm's law gives the source current: I = 18 / 6 = 3 A. All of this current crosses the 4 Ω resistor, so its voltage drop is 3 × 4 = 12 V. That leaves 18 − 12 = 6 V between A and B.
Each parallel branch has that same 6 V across it. The branch currents are 6 / 6 = 1 A and 6 / 3 = 2 A. Their sum, 3 A, matches the current entering the junction.
| Resistor | Voltage | Current | Power |
|---|---|---|---|
| 4 Ω, series | 12 V | 3 A | 36 W |
| 6 Ω, branch | 6 V | 1 A | 6 W |
| 3 Ω, branch | 6 V | 2 A | 12 W |
4. Check the result in a different way
The source supplies P = VI = 18 × 3 = 54 W. The resistors dissipate 36 + 6 + 12 = 54 W. Matching totals support the calculation. A mismatch points to an incorrect current, voltage or arithmetic step; it is a reason to revisit the working, not to force the final number to match.
Keep units beside each answer. For example, 18 / 6 here is volts divided by ohms, so the answer is amperes. When a problem uses kΩ or mA, convert consistently before applying the same steps.
Try it before opening the answers
A. Change only the source
Use the same resistors with a 12 V source. Find the source current and both branch currents.
Show the working
The resistance stays 6 Ω, so the source current is 2 A. The 4 Ω resistor drops 8 V, leaving 4 V across the parallel pair. Branch currents are 4 / 6 = 2/3 A and 4 / 3 A. They add to 2 A. Every voltage and current is two-thirds of its value in the 18 V example.
B. Diagnose a wrong answer
A student uses 18 / 3 and reports 6 A in the 3 Ω branch. Which assumption is wrong?
Show the explanation
The student applied the full source voltage to one parallel branch. The series resistor also has a voltage drop. Calculate the voltage between the shared branch endpoints first; it is 6 V in the original example, giving 2 A in the 3 Ω branch.
What to try next
Change one resistance and repeat all four stages: connections, equivalent resistance, branch values and power check. If a network cannot be reduced into series and parallel groups, your next topic is Kirchhoff's laws and systematic node or loop equations.
Further explanation: OpenStax, University Physics Volume 2, section 10.2. The worked values and practice questions above were prepared for this guide.