Series vs parallel circuits, with worked examples

In a series circuit, every resistor carries the same current and the voltage is shared. In a parallel circuit, every resistor has the same voltage and the current is shared. Almost every resistor-network question in College Physics 2 comes down to knowing which of those two statements applies, then using V = IR.

The two rules

Series (one path, end to end):

• Equivalent resistance: R_s = R1 + R2 + R3 + ...
• The same current I flows through each resistor.
• The voltage drops add up to the source voltage: V = V1 + V2 + V3 + ...
• R_s is always larger than the largest single resistor.

Parallel (each resistor on its own branch between the same two points):

• Equivalent resistance: 1/R_p = 1/R1 + 1/R2 + 1/R3 + ...
• The same voltage V is across each resistor.
• The branch currents add up to the total current: I = I1 + I2 + I3 + ...
• R_p is always smaller than the smallest single resistor.

For exactly two resistors in parallel there is a shortcut: R_p = R1 x R2 / (R1 + R2). For n identical resistors R in parallel, R_p = R / n.

The "always larger" and "always smaller" facts are the fastest way to catch a wrong answer. If you add a 3 ohm and a 6 ohm resistor in parallel and get anything above 3 ohm, something went wrong.

Example 1: three resistors in series

A 12.0 V battery drives 1.00, 6.00 and 13.0 ohm resistors in series. This is the setup of OpenStax College Physics Example 21.1. Assume the battery has no internal resistance.

• R_s = 1.00 + 6.00 + 13.0 = 20.0 ohm
• I = V / R_s = 12.0 / 20.0 = 0.600 A, the same through every resistor
• Drops (V = IR): 0.600 x 1.00 = 0.600 V, 0.600 x 6.00 = 3.60 V, 0.600 x 13.0 = 7.80 V
• Check: 0.600 + 3.60 + 7.80 = 12.0 V

Power in each resistor is P = I^2 R: 0.360 W, 2.16 W and 4.68 W. They add to 7.20 W, which equals the power the battery delivers, P = IV = 12.0 x 0.600 = 7.20 W. The largest resistor in a series string gets the largest share of the voltage and dissipates the most power.

Example 2: the same resistors in parallel

Now connect the same three resistors in parallel across the same 12.0 V battery (OpenStax Example 21.2).

• 1/R_p = 1/1.00 + 1/6.00 + 1/13.0 = 1.244 per ohm, so R_p = 0.804 ohm
• Each resistor has the full 12.0 V across it.
• Currents (I = V/R): 12.0 / 1.00 = 12.0 A, 12.0 / 6.00 = 2.00 A, 12.0 / 13.0 = 0.923 A
• Total current: 12.0 + 2.00 + 0.923 = 14.9 A, which matches 12.0 / 0.804

The powers are now 144 W, 24.0 W and 11.1 W, about 179 W in all. Same resistors, same battery, roughly 25 times the power of the series version. That is why household outlets are wired in parallel: every appliance gets the full voltage, and each device you add raises the total current the circuit carries. It is also why too many devices on one circuit trips a breaker.

In parallel it is the smallest resistor that carries the most current and uses the most power, the reverse of series.

Example 3: two quick parallel checks

• Two 100 ohm resistors in parallel: 100 / 2 = 50 ohm. Identical resistors in parallel always give R / n.
• 6.00 ohm and 3.00 ohm in parallel: (6.00 x 3.00) / (6.00 + 3.00) = 18.0 / 9.00 = 2.00 ohm. With 12.0 V across the pair, the 6 ohm branch takes 2.00 A and the 3 ohm branch takes 4.00 A, for 6.00 A total.

The current splits in inverse proportion to resistance: the 3 ohm branch has half the resistance, so it carries twice the current.

Example 4: a mixed circuit

Real problems often combine the two. Put a 4.00 ohm resistor in series with the 6.00 ohm and 3.00 ohm parallel pair from Example 3, and connect a 12.0 V source.

1. Reduce the parallel part first: 6.00 || 3.00 = 2.00 ohm.
2. That 2.00 ohm is now in series with the 4.00 ohm: R_eq = 4.00 + 2.00 = 6.00 ohm.
3. Total current: I = 12.0 / 6.00 = 2.00 A. All of it goes through the 4.00 ohm resistor.
4. Drop across the 4.00 ohm: 2.00 x 4.00 = 8.00 V. That leaves 12.0 - 8.00 = 4.00 V across the parallel pair (check: 2.00 A x 2.00 ohm = 4.00 V).
5. Branch currents: 4.00 / 6.00 = 0.667 A and 4.00 / 3.00 = 1.33 A. They add back to 2.00 A.

The method is always the same: collapse the innermost series or parallel groups, work out the total current, then expand back out one step at a time. When a network cannot be reduced this way (for example, two batteries in different branches), you need Kirchhoff's rules instead, which come later in the same unit.

Mistakes that cost points

• Forgetting to invert. 1/R_p = 1/6 + 1/3 = 0.5 is not the answer. R_p = 1 / 0.5 = 2 ohm.
• Using the two-resistor shortcut on three resistors. R1R2/(R1 + R2) only works for a pair. For three, use the reciprocal sum or apply the shortcut twice.
• Mixing up the capacitor rules. Capacitors combine the opposite way: they add directly in parallel and by reciprocals in series.
• Assuming the current is the same everywhere. It is only the same through elements in series. In Example 4, the 4 ohm resistor carries 2.00 A but neither parallel branch does.
• Unit prefixes. A 4.7 kilohm resistor is 4,700 ohm. Convert before you divide.

Check your answers

The series and parallel resistor calculator gives the equivalent resistance for up to 8 resistors and, with a source voltage, the current, voltage drop and power for each one. Use it to check work, not replace it: exams want the reduction steps. The battery's voltage itself comes from chemistry, covered in the electrochemistry unit of General Chemistry II. For where circuits sit in the course, see what's in College Physics 2; for the force law that underlies it all, see Coulomb's law and electric fields, worked examples. How to study for a physics exam covers practice strategy, and how to propagate uncertainty in a lab report helps when a circuits lab asks for error analysis.

Encodr's free College Physics 2 course, with cards on every circuit rule here, is coming soon; the current courses are on the decks page.