Series vs Parallel Circuits: Equivalent Resistance

Series resistors share one path, so the same current flows through each and resistances add: Req = R1 + R2. Parallel resistors share one voltage (the full battery voltage when directly across it), and reciprocals add, so Req is smaller than the smallest branch. Reduce to one Req and use I = V/R.

AP Physics: Unit 11 (topics 11.2 Simple Circuits, 11.5 Compound Direct Current (DC) Circuits). This page covers Topic 11.5, Compound Direct Current (DC) Circuits, in AP Physics 2 Unit 11 (Electric Circuits, weighted 15 to 18 percent). AP Physics C: Electricity and Magnetism tests the same reductions in its own Unit 11 (Electric Circuits, weighted 15 to 25 percent).

Series vs parallel: what actually changes

Series and parallel are the two ways to wire resistors, and they behave in opposite ways. In series, resistors connect end to end, so charge has exactly one path from the battery's positive terminal back to its negative terminal. Every extra series resistor makes that single path longer and harder to get through, so the equivalent resistance rises. In parallel, resistors sit side by side between the same two junctions. Each new branch gives charge another route between those junctions, so the equivalent resistance drops even though you added a resistor.

The quantity that stays the same also flips. Series elements share one current, because charge has nowhere else to go. Parallel branches share one voltage, because both ends of every branch connect to the same two points. Almost every circuits question comes down to applying those two facts in the right order.

Series: one path, one current

Wire two resistors in series and the same current passes through both. Current is a flow of charge, and charge does not pile up or vanish in a steady-state circuit, so whatever enters $R_1$ each second must leave $R_2$ each second. What divides is the voltage: the battery's potential difference splits among the resistors in proportion to their resistance, since $\Delta V = IR$ with the same $I$ everywhere (Ohm's law does most of the work here).

The equivalent resistance is a straight sum:

R_{eq} = R_1 + R_2 + R_3 + \dots

The sum is always bigger than any individual resistor. A useful mental model: two resistors in series act like one longer resistor, and a longer path of the same material always has more resistance.

Parallel: one voltage, current splits

Wire two resistors in parallel and each one feels the full potential difference between the two shared junctions. Voltage is energy per coulomb between two points, and both branches span the same two points, so both get the same $\Delta V$ . What splits is the current: at the junction, more charge per second flows through the smaller resistance, since $I = \Delta V / R$ with the same $\Delta V$ for each branch.

Resistances combine through reciprocals:

\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots

The result is always smaller than the smallest branch, because every added branch is an extra lane for charge. Two shortcuts worth memorizing: for exactly two resistors, $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$ (product over sum), and $N$ identical resistors of value $R$ in parallel give $R_{eq} = R/N$ .

The rules side by side

This table settles most conceptual questions about compound circuits before you touch a calculator, so it is worth knowing cold.

Quantity	Series	Parallel
Current	Same through every element	Splits among branches
Voltage	Divides across elements	Same across every branch
Equivalent resistance	$R_{eq} = \sum R_i$	$1/R_{eq} = \sum 1/R_i$
Adding a resistor	$R_{eq}$ increases	$R_{eq}$ decreases
Picture	Longer path	More lanes

One warning about the parallel rule: after summing the reciprocals, you must flip the result to get $R_{eq}$ . Skipping that final flip is an easy slip, and also an easy one to catch, because a correct parallel $R_{eq}$ is always smaller than the smallest resistor in the group. If your answer is not, flip it.

How to find current in a circuit

Finding the current anywhere in a resistor network is a two-pass process. First collapse the network to a single equivalent resistance and get the battery current from Ohm's law. Then expand the circuit back out one step at a time, carrying what you know at each stage.

Reduce: combine series chains and parallel groups until one $R_{eq}$ remains.
Solve: $I = \Delta V / R_{eq}$ gives the current through the battery.
Expand: reverse each combination in the opposite order. Series elements inherit the current of the combination they came from; parallel branches inherit its voltage. Apply $I = \Delta V / R$ to each resistor as its values appear.

Label every intermediate current and voltage as you go instead of holding them in your head. You can check any single step with the Ohm's law calculator, which solves $I = \Delta V / R$ for whichever variable is missing.

Reducing a mixed circuit

AP problems rarely hand you a pure series or pure parallel circuit. The standard setup is a compound circuit, such as one resistor in series with a parallel pair. Reduce from the inside out: find the innermost group whose members are cleanly in series or parallel, replace it with one equivalent resistor, redraw the circuit, and repeat until a single resistor remains.

Use these tests before combining anything. Two resistors are in series only if they share a junction with nothing else attached to it, so the same current must pass through both. Two resistors are in parallel only if both of their ends connect to the same two junctions. If neither test passes, that pair cannot be combined yet; reduce a different part of the circuit first and check again. Some networks never reduce, such as a resistor bridging the middle of two branches; those call for Kirchhoff's loop and junction rules, Topics 11.6 and 11.7 in AP Physics 2. The third worked example below runs a full reduction, then unwinds it.

Mistakes to catch before the exam

Summing reciprocals and forgetting to flip: $6.0\ \Omega$ and $3.0\ \Omega$ in parallel make $2.0\ \Omega$ , not $0.50\ \Omega$ .
Assuming current splits evenly between parallel branches. It splits in inverse proportion to resistance, so the smaller resistor carries more of the current.
Calling resistors in series because they look lined up in the diagram. If a junction between them leads anywhere else, they are not in series.
Applying the battery's full voltage to a resistor buried inside a compound circuit. Only elements connected directly across the battery get the full $\Delta V$ ; everything else gets its share, which you find during the expansion pass.
Mixing up which quantity is constant. Same current in series, same voltage in parallel. If you remember one line from this page, make it that one.

Series circuit: current and voltage drops

A 12.0 V battery is connected to a $3.0\ \Omega$ resistor and a $5.0\ \Omega$ resistor in series. Find the equivalent resistance, the current, and the voltage across each resistor.

Add the resistances. In series, $R_{eq} = R_1 + R_2 = 3.0\ \Omega + 5.0\ \Omega = 8.0\ \Omega$ .
Apply Ohm's law to the reduced circuit. $I = \Delta V / R_{eq} = 12.0\ \text{V} / 8.0\ \Omega = 1.5\ \text{A}$ . The same 1.5 A flows through both resistors, because a series circuit has only one path.
Find each voltage drop. $\Delta V_1 = I R_1 = (1.5\ \text{A})(3.0\ \Omega) = 4.5\ \text{V}$ and $\Delta V_2 = I R_2 = (1.5\ \text{A})(5.0\ \Omega) = 7.5\ \text{V}$ .
Check. $4.5\ \text{V} + 7.5\ \text{V} = 12.0\ \text{V}$ , the full battery voltage. The larger resistor takes the larger share of the voltage.

$R_{eq} = 8.0\ \Omega$ and $I = 1.5\ \text{A}$ , with 4.5 V across the $3.0\ \Omega$ resistor and 7.5 V across the $5.0\ \Omega$ resistor.

Parallel circuit: equivalent resistance and branch currents

A 12.0 V battery is connected across a $6.0\ \Omega$ resistor and a $3.0\ \Omega$ resistor in parallel. Find the equivalent resistance, the total current from the battery, and the current in each branch.

Sum the reciprocals. $1/R_{eq} = 1/(6.0\ \Omega) + 1/(3.0\ \Omega) = 0.167\ \Omega^{-1} + 0.333\ \Omega^{-1} = 0.500\ \Omega^{-1}$ .
Flip to get the equivalent resistance. $R_{eq} = 1 / (0.500\ \Omega^{-1}) = 2.0\ \Omega$ . Sanity check: 2.0 is smaller than 3.0, the smallest branch, as a parallel combination must be.
Find the battery current. $I = \Delta V / R_{eq} = 12.0\ \text{V} / 2.0\ \Omega = 6.0\ \text{A}$ .
Split the current by branch. Each branch gets the full 12.0 V: $I_1 = 12.0\ \text{V} / 6.0\ \Omega = 2.0\ \text{A}$ and $I_2 = 12.0\ \text{V} / 3.0\ \Omega = 4.0\ \text{A}$ .
Check. $2.0\ \text{A} + 4.0\ \text{A} = 6.0\ \text{A}$ , matching the battery current. The smaller resistor carries twice the current, as the inverse proportion predicts.

$R_{eq} = 2.0\ \Omega$ ; the battery supplies 6.0 A, with 2.0 A through the $6.0\ \Omega$ branch and 4.0 A through the $3.0\ \Omega$ branch.

Mixed circuit: reduce, then unwind

A 24.0 V battery drives a $4.0\ \Omega$ resistor in series with a parallel pair, $6.0\ \Omega$ and $12.0\ \Omega$ . Find the current through and the voltage across every resistor.

Combine the parallel pair first. $1/R_p = 1/(6.0\ \Omega) + 1/(12.0\ \Omega) = 0.167\ \Omega^{-1} + 0.0833\ \Omega^{-1} = 0.250\ \Omega^{-1}$ , so $R_p = 4.0\ \Omega$ .
Combine the series chain. The $4.0\ \Omega$ resistor is in series with the $4.0\ \Omega$ equivalent: $R_{eq} = 4.0\ \Omega + 4.0\ \Omega = 8.0\ \Omega$ .
Find the battery current. $I = \Delta V / R_{eq} = 24.0\ \text{V} / 8.0\ \Omega = 3.0\ \text{A}$ . All 3.0 A passes through the series $4.0\ \Omega$ resistor, since it sits on the single main path.
Unwind the series step. Voltage across the series resistor: $\Delta V_1 = (3.0\ \text{A})(4.0\ \Omega) = 12.0\ \text{V}$ . The parallel pair gets the remainder: $24.0\ \text{V} - 12.0\ \text{V} = 12.0\ \text{V}$ .
Unwind the parallel step. Both branches see 12.0 V: $I_6 = 12.0\ \text{V} / 6.0\ \Omega = 2.0\ \text{A}$ and $I_{12} = 12.0\ \text{V} / 12.0\ \Omega = 1.0\ \text{A}$ . Check: $2.0\ \text{A} + 1.0\ \text{A} = 3.0\ \text{A}$ , matching the battery current.

The battery supplies 3.0 A. The $4.0\ \Omega$ resistor carries 3.0 A with 12.0 V across it; the $6.0\ \Omega$ branch carries 2.0 A and the $12.0\ \Omega$ branch carries 1.0 A, each with 12.0 V across it.

Frequently asked questions

Why is equivalent resistance smaller in parallel?

Every parallel branch gives charge another path between the same two points. More paths means more total current at the same voltage, and since equivalent resistance is voltage divided by total current, more current means less resistance. Adding a branch always pushes Req below the smallest resistor in the group, no matter how large the new resistor is.

What stays constant in series and what stays constant in parallel?

Current is the same through every element in series, because there is only one path for charge. Voltage is the same across every branch in parallel, because all branches connect the same two junctions. The other quantity divides: voltage splits across series resistors, and current splits among parallel branches.

How do I find the current in a circuit with both series and parallel parts?

Reduce the circuit from the inside out until one equivalent resistance remains, then use Ohm's law (I = V/R) to get the battery current. Next, work backward through your reductions: series pieces keep the current of the combination they came from, and parallel pieces keep its voltage. The mixed-circuit worked example on this page shows the full process.

Are the series and parallel resistance formulas on the AP equation sheet?

Yes. The AP Physics 2 equation sheet lists the series rule (resistances add) and the parallel rule (reciprocals add), along with Ohm's law in the form I = V/R. You still have to remember the last move in a parallel calculation yourself: flip the reciprocal sum to get Req.

Why does adding a resistor in parallel increase the total current from the battery?

The new branch does not disturb the old ones: each existing branch still has the same voltage across it, so each still draws its original current. The new branch draws extra current on top of that, so the battery supplies more in total. That is exactly what a drop in equivalent resistance means.