# Ohm's law and simple circuits

Canonical: https://duckyhelper.com/learn/physics/ohms-law-and-circuits/
Updated: 2026-10-01

Ohm's law says the current through a resistor equals the voltage across it divided by its resistance: \(V = IR\). Voltage (volts) is the push, current (amperes) is the flow of charge, and resistance (ohms) is how hard the part makes that flow. In a series circuit, resistances add and the current is the same everywhere. In a parallel circuit, each branch gets the same voltage and the branch currents add up.

## Key ideas

$$
V = I R
$$

- \(V\) is the voltage (potential difference) across the part, in volts (V).
- \(I\) is the current through it, in amperes (A). 1 A is 1 coulomb of charge per second.
- \(R\) is the resistance, in ohms (\(\Omega\)).

$$
R_{\text{series}} = R_1 + R_2 + \cdots \qquad \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots
$$

$$
P = IV = I^2 R = \frac{V^2}{R}
$$

\(P\) is electric power, in watts (W): the energy used per second. Your power company bills energy in kilowatt-hours (kWh), which is power times time.

**Series vs parallel**

|  | Series | Parallel |
| --- | --- | --- |
| Layout | One path, parts in a line | Several branches side by side |
| Current | Same through every part | Splits; branch currents add to the total |
| Voltage | Splits; the drops add to the battery voltage | Same across every branch |
| Total resistance | Sum, bigger than any one resistor | Smaller than the smallest resistor |
| If one bulb breaks | Everything goes out | The other branches stay on |

## Worked examples

**Example 1: one resistor**

Problem: A 9.0 V battery is connected to a 450 \(\Omega\) resistor. What current flows?

1. Solve Ohm's law for current.

   $$
   I = \frac{V}{R} = \frac{9.0\ \text{V}}{450\ \Omega} = 0.020\ \text{A}
   $$
2. 0.020 A is 20 milliamperes (mA).

Answer: 0.020 A (20 mA)

**Example 2: two resistors in series**

Problem: A 12 V battery runs a 4.0 \(\Omega\) and an 8.0 \(\Omega\) resistor in series. Find the current and the voltage across each resistor.

1. Total resistance.

   $$
   R = 4.0 + 8.0 = 12\ \Omega
   $$
2. Current, the same everywhere in series.

   $$
   I = \frac{12\ \text{V}}{12\ \Omega} = 1.0\ \text{A}
   $$
3. Voltage across each.

   $$
   V_1 = (1.0\ \text{A})(4.0\ \Omega) = 4.0\ \text{V} \qquad V_2 = (1.0\ \text{A})(8.0\ \Omega) = 8.0\ \text{V}
   $$
4. Check: \(4.0 + 8.0 = 12\) V, the battery voltage.

Answer: 1.0 A; 4.0 V and 8.0 V

**Example 3: two resistors in parallel**

Problem: A 12 V battery runs a 6.0 \(\Omega\) and a 3.0 \(\Omega\) resistor in parallel. Find the total resistance, the total current and the current in each branch.

1. Total resistance.

   $$
   \frac{1}{R} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{3}{6.0} \quad\Rightarrow\quad R = 2.0\ \Omega
   $$
2. Total current.

   $$
   I = \frac{12\ \text{V}}{2.0\ \Omega} = 6.0\ \text{A}
   $$
3. Each branch has the full 12 V.

   $$
   I_1 = \frac{12}{6.0} = 2.0\ \text{A} \qquad I_2 = \frac{12}{3.0} = 4.0\ \text{A}
   $$
4. Check: \(2.0 + 4.0 = 6.0\) A.

Answer: 2.0 Ω total, 6.0 A total; 2.0 A and 4.0 A in the branches

**Example 4: power of a light bulb**

Problem: A 60. W bulb runs on 120 V. What current does it draw, and what is its resistance while lit?

1. Current from \(P = IV\).

   $$
   I = \frac{P}{V} = \frac{60.\ \text{W}}{120\ \text{V}} = 0.50\ \text{A}
   $$
2. Resistance from Ohm's law.

   $$
   R = \frac{V}{I} = \frac{120\ \text{V}}{0.50\ \text{A}} = 240\ \Omega
   $$

Answer: 0.50 A and 240 Ω

## Common mistakes and how to fix them

- **Adding parallel resistors like series ones.** Fix: in parallel, add the reciprocals, then flip the answer. Two equal resistors in parallel give half of one.
- **Forgetting the last flip.** \(\tfrac{1}{R} = 0.5\) means \(R = 2.0\ \Omega\), not 0.5. Fix: the parallel total must be smaller than the smallest resistor.
- **Using the battery voltage across one series resistor.** In series, the voltage is shared. Fix: find the current first, then \(V = IR\) for each part.
- **Mixing milliamps and amps.** Fix: 1 mA = 0.001 A. Convert before using \(V = IR\).

**Practice problems**

1. A 6.0 V battery is connected to a 3.0 \(\Omega\) resistor. What is the current?
   A. 0.50 A
   B. 2.0 A
   C. 18 A
   D. 9.0 A

   Answer: 2.0 A. \(I = V/R = 6.0 \div 3.0 = 2.0\) A.

2. What is the total resistance of 10. \(\Omega\), 20. \(\Omega\) and 30. \(\Omega\) resistors in series?

   Answer: 60. Ω. In series, resistances add: \(10. + 20. + 30. = 60.\ \Omega\).

3. What is the total resistance of two 10. \(\Omega\) resistors in parallel?
   A. 20. Ω
   B. 10. Ω
   C. 5.0 Ω
   D. 0.20 Ω

   Answer: 5.0 Ω. \(\tfrac{1}{R} = \tfrac{1}{10.} + \tfrac{1}{10.} = 0.20\), so \(R = 5.0\ \Omega\). Two equal resistors in parallel give half of one.

4. In which kind of circuit is the current the same through every part?
   A. Series
   B. Parallel

   Answer: Series. A series circuit has only one path, so every bit of charge goes through every part.

5. A 1800 W space heater runs on 120 V. What current does it draw?

   Answer: 15 A. \(I = P/V = 1800 \div 120 = 15\) A. That is a lot for one outlet, which is why heaters can trip a circuit breaker.

## Frequently asked questions

### What is the difference between voltage and current?

A water pipe is a good picture. Voltage is like the water pressure that pushes, and current is like how much water flows past a point each second. Resistance is like a narrow section of pipe. More pressure or a wider pipe gives more flow.

### Why are houses wired in parallel?

In parallel, every outlet gets the full voltage, and each device can be switched on or off without affecting the others. If houses were wired in series, turning off one lamp would turn off everything, and the voltage would be shared between devices.

### Does Ohm's law work for everything?

No. It works for ohmic parts like ordinary resistors, where resistance stays constant. A light bulb filament's resistance rises as it heats up, and diodes and LEDs do not follow \(V = IR\) with a constant R. For high school circuit problems you can assume ohmic resistors.

## Sources

- [OpenStax College Physics 2e, 20.2 Ohm's Law: Resistance and Simple Circuits](https://openstax.org/books/college-physics-2e/pages/20-2-ohms-law-resistance-and-simple-circuits), accessed 2026-10-01
- [OpenStax College Physics 2e, 20.4 Electric Power and Energy](https://openstax.org/books/college-physics-2e/pages/20-4-electric-power-and-energy), accessed 2026-10-01
- [OpenStax College Physics 2e, 21.1 Resistors in Series and Parallel](https://openstax.org/books/college-physics-2e/pages/21-1-resistors-in-series-and-parallel), accessed 2026-10-01

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## Try asking Ducky

- "Check my parallel resistance on number 3. I think I forgot to flip the fraction."
- "Look at my circuit drawing. Which resistors are in series and which are in parallel?"
- "Give me a circuit with one series and two parallel resistors to solve step by step."

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