Ohm's Law Calculator + Series/Parallel Resistance

Enter any two of voltage, current, resistance, and power, and the calculator above solves the other two using Ohm's law (I = V/R) and P = IV. In resistance mode, it combines any list of resistors: values add in series, and reciprocals add in parallel.

Voltage V (V)Current I (A)Resistance R (ohms)Power P (W)Computed output; leave this one empty.

solved from V and I

R = 6 ohms, P = 24 W

Full set: V = 12 V, I = 2 A, R = 6 ohms, P = 24 W.

Steps

1.given: V = 12 V and I = 2 A
2.R = V / I = 12 V / 2 A = 6 ohms
3.P = V I = 12 V x 2 A = 24 W

AP Physics: Unit 11 (topics 11.3 Resistance, Resistivity, and Ohm's Law, 11.5 Compound Direct Current (DC) Circuits). Built for AP Physics 2 Unit 11 (Electric Circuits), specifically Topic 11.3 on Ohm's law and Topic 11.5 on compound DC circuits. The same series and parallel reduction rules also appear in AP Physics C: Electricity and Magnetism Unit 11.

What the calculator above solves

The calculator above does two jobs. In Ohm's law mode, enter any two of potential difference $\Delta V$ (volts), current $I$ (amperes), resistance $R$ (ohms), and power $P$ (watts), and it solves for the other two. In resistance mode, type in a list of resistor values and it returns the equivalent resistance for a series arrangement and a parallel arrangement of those same resistors.

Two inputs are enough because the four quantities are tied together by just two independent equations: Ohm's law and the power equation. Every other formula you have seen, like $P = I^2 R$ , is a combination of those two. If your numbers come from an AP-style problem, keep everything in volts, amperes, and ohms so the outputs land in watts with no conversion step.

Ohm's law, as the AP sheet writes it

The AP Physics 2 equation sheet prints Ohm's law as $I = \frac{\Delta V}{R}$ : the current through a resistor equals the potential difference across it divided by its resistance. Rearranged, $\Delta V = IR$ and $R = \Delta V / I$ . Doubling the voltage across a fixed resistor doubles the current, and doubling the resistance at fixed voltage halves the current.

A quick unit check keeps you honest. One ampere is one volt per ohm, so 12 V across 30 Ω can only give 0.40 A, never 360 A. If your answer explodes, you probably multiplied where you should have divided. For the physics behind the equation, including what counts as an ohmic material and how resistivity and geometry set resistance, read the full Ohm's law guide.

Power: the second equation the solver uses

Electric power measures how fast electrical energy is converted in a circuit element, and the AP sheet prints it as $P = I \Delta V$ . Combine it with Ohm's law and you get two more forms:

P = I^2 R = \frac{(\Delta V)^2}{R}

These substitutions are why any two inputs pin down all four values. Given power and resistance, the calculator solves $P = I^2 R$ for the current, $I = \sqrt{P/R}$ , then finds the voltage from $\Delta V = IR$ . Try it: a resistor dissipating 5.0 W with a resistance of 20 Ω carries $I = \sqrt{5.0/20} = 0.50$ A and sits across $\Delta V = (0.50)(20) = 10$ V. Watch the units, though. Power comes out in watts only when you feed in volts, amperes, and ohms, so if a problem hands you kilohms, convert first: 2.2 kΩ is 2200 Ω.

Series resistors: values add

Resistors in series sit along a single path, so the same current passes through each one and their potential differences add. The equivalent resistance is the plain sum:

R_{eq} = R_1 + R_2 + \dots

Two-resistor example: a 40 Ω and a 60 Ω resistor in series give $R_{eq} = 40 + 60 = 100$ Ω. Across a 10 V battery the current is $I = 10/100 = 0.10$ A everywhere in the loop. The 40 Ω resistor takes $\Delta V = (0.10)(40) = 4.0$ V and the 60 Ω resistor takes 6.0 V, and those two voltages add back to the full 10 V, a check worth writing out on the exam. Adding a resistor in series always increases the equivalent resistance. For how to tell whether a real circuit counts as series in the first place, see series vs parallel circuits.

Parallel resistors: reciprocals add

Resistors in parallel connect the same two nodes, so each one sees the same potential difference and the branch currents add. For resistance, the reciprocals add:

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

Two-resistor example: 40 Ω and 60 Ω in parallel give $1/R_{eq} = 1/40 + 1/60 = 3/120 + 2/120 = 5/120$ , so $R_{eq} = 120/5 = 24$ Ω. Notice that 24 Ω is smaller than either resistor. Adding a branch opens another path for current, so the combination conducts better than any single branch. For exactly two resistors you can shortcut with product over sum: $R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{(40)(60)}{100} = 24$ Ω. The classic mistake is stopping at 5/120 and reporting that as the answer, when it is actually 1 over the answer. The calculator above does the final reciprocal for you, but write it explicitly in your own work.

Where this lands on the AP exams

This tool lines up with AP Physics 2 Unit 11, Electric Circuits, which carries a 15-18% exam weighting. Topic 11.3 covers resistance, resistivity, and Ohm's law, and Topic 11.5 covers compound DC circuits, where you reduce series and parallel combinations to one equivalent resistance before solving anything else. The same rules return in AP Physics C: Electricity and Magnetism, where Unit 11, Electric Circuits, is weighted at 15-25%.

The exam itself runs 3 hours: 42 multiple-choice questions in 85 minutes, then 4 free-response questions in 95 minutes, with each section worth 50% of your score. A four-function, scientific, or graphing calculator is allowed on both sections, so a good study loop is to reduce a resistor network by hand, then confirm with the tool above. The full printed equation list lives on the AP Physics 2 formula sheet page.

Current and power from voltage and resistance

A 12.0 V battery is connected across a single 30.0 Ω resistor. Find the current through the resistor and the power it dissipates.

Solve Ohm's law for current: $I = \Delta V / R = 12.0\ \text{V} / 30.0\ \Omega = 0.400\ \text{A}$ .
Use the power equation from the AP sheet: $P = I \Delta V = (0.400\ \text{A})(12.0\ \text{V}) = 4.80\ \text{W}$ .
Check with the substituted form: $P = (\Delta V)^2 / R = (12.0\ \text{V})^2 / 30.0\ \Omega = 144 / 30.0 = 4.80\ \text{W}$ . Both routes agree.

$I = 0.400$ A and $P = 4.80$ W.

Three resistors in series

A 10.0 Ω, a 22.0 Ω, and a 47.0 Ω resistor are connected in series with a 9.00 V battery. Find the equivalent resistance, the current, and the potential difference across the 47.0 Ω resistor.

Series values add: $R_{eq} = 10.0 + 22.0 + 47.0 = 79.0\ \Omega$ .
One current flows through the whole loop: $I = \Delta V / R_{eq} = 9.00\ \text{V} / 79.0\ \Omega = 0.114\ \text{A}$ .
Voltage across the 47.0 Ω resistor. Carry the unrounded current 0.1139 A so the drop stays consistent: $\Delta V_{47} = I R = (0.1139\ \text{A})(47.0\ \Omega) = 5.35\ \text{V}$ .
Check: the other drops are $(0.114)(10.0) = 1.14$ V and $(0.114)(22.0) = 2.51$ V, and $1.14 + 2.51 + 5.35 = 9.00$ V, matching the battery.

$R_{eq} = 79.0$ Ω, $I = 0.114$ A, and the 47.0 Ω resistor takes 5.35 V.

Two resistors in parallel

An 8.00 Ω resistor and a 12.0 Ω resistor are connected in parallel across a 24.0 V source. Find the equivalent resistance, the total current, and each branch current.

Add reciprocals, keeping an extra digit: $1/R_{eq} = 1/8.00 + 1/12.0 = 0.125 + 0.0833 = 0.2083\ \Omega^{-1}$ .
Flip to get the equivalent resistance: $R_{eq} = 1 / 0.2083 = 4.80\ \Omega$ . As expected, 4.80 Ω is smaller than either branch.
Total current from the source: $I = \Delta V / R_{eq} = 24.0\ \text{V} / 4.80\ \Omega = 5.00\ \text{A}$ .
Each branch sees the full 24.0 V: $I_8 = 24.0 / 8.00 = 3.00\ \text{A}$ and $I_{12} = 24.0 / 12.0 = 2.00\ \text{A}$ .
Check: $3.00 + 2.00 = 5.00$ A, matching the total from step 3.

$R_{eq} = 4.80$ Ω, total current 5.00 A, with 3.00 A through the 8.00 Ω branch and 2.00 A through the 12.0 Ω branch.

Frequently asked questions

Which two values do I need to enter into the calculator?

Any two of the four: voltage, current, resistance, or power. Ohm's law (I = V/R) plus the power equation (P = IV) form a system of two equations, so two known values always determine the other two.

Why is the parallel equivalent resistance smaller than the smallest resistor?

Every parallel branch is an extra path for current. Adding a path can only make it easier for charge to flow at a given voltage, which means more total current and therefore a lower equivalent resistance. With 40 Ω and 60 Ω in parallel, the result is 24 Ω, below both.

Does Ohm's law apply to every circuit element?

No. It describes ohmic materials, where resistance stays constant as voltage changes. A light bulb filament heats up as current increases, so its resistance rises and its current-voltage graph curves. You can still compute R = V/I at any single operating point, but that R is not a fixed constant for the device.

What stays the same in series, and what stays the same in parallel?

In series, the current is identical through every resistor and the voltages add up to the battery voltage. In parallel, the voltage is identical across every branch and the branch currents add up to the total. Mixing those two facts up is the fastest way to a wrong answer on compound circuits, so check the series vs parallel guide if you are unsure.

Can I use a calculator on the AP Physics 2 exam?

Yes. A four-function, scientific, or graphing calculator is allowed on both the multiple-choice section and the free-response section, so you can crunch equivalent resistance and power numbers directly on exam day.