Ohm's law and circuits: series, parallel and power

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?

Solve Ohm's law for current.
$I = \frac{V}{R} = \frac{9.0\ \text{V}}{450\ \Omega} = 0.020\ \text{A}$
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.

Total resistance.
$R = 4.0 + 8.0 = 12\ \Omega$
Current, the same everywhere in series.
$I = \frac{12\ \text{V}}{12\ \Omega} = 1.0\ \text{A}$
Voltage across each.
$V_1 = (1.0\ \text{A})(4.0\ \Omega) = 4.0\ \text{V} \qquad V_2 = (1.0\ \text{A})(8.0\ \Omega) = 8.0\ \text{V}$
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.

Total resistance.
$\frac{1}{R} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{3}{6.0} \quad\Rightarrow\quad R = 2.0\ \Omega$
Total current.
$I = \frac{12\ \text{V}}{2.0\ \Omega} = 6.0\ \text{A}$
Each branch has the full 12 V.
$I_1 = \frac{12}{6.0} = 2.0\ \text{A} \qquad I_2 = \frac{12}{3.0} = 4.0\ \text{A}$
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?

Current from $P = IV$ .
$I = \frac{P}{V} = \frac{60.\ \text{W}}{120\ \text{V}} = 0.50\ \text{A}$
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

A 6.0 V battery is connected to a 3.0 $\Omega$ resistor. What is the current?
1. 0.50 A
2. 2.0 A
3. 18 A
4. 9.0 A
Show answer
Answer: 2.0 A
$I = V/R = 6.0 \div 3.0 = 2.0$ A.
What is the total resistance of 10. $\Omega$ , 20. $\Omega$ and 30. $\Omega$ resistors in series?
Show answer
Answer: 60. Ω
In series, resistances add: $10. + 20. + 30. = 60.\ \Omega$ .
What is the total resistance of two 10. $\Omega$ resistors in parallel?
1. 20. Ω
2. 10. Ω
3. 5.0 Ω
4. 0.20 Ω
Show answer
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.
In which kind of circuit is the current the same through every part?
1. Series
2. Parallel
Show answer
Answer: Series
A series circuit has only one path, so every bit of charge goes through every part.
A 1800 W space heater runs on 120 V. What current does it draw?
Show answer
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, accessed October 1, 2026
OpenStax College Physics 2e, 20.4 Electric Power and Energy, accessed October 1, 2026
OpenStax College Physics 2e, 21.1 Resistors in Series and Parallel, accessed October 1, 2026

Ohm's law and simple circuits

Key ideas

Worked examples

Common mistakes and how to fix them

Frequently asked questions

Sources

Try asking Ducky

Key ideas

Worked examples

Common mistakes and how to fix them

Frequently asked questions

Sources

Related

Try asking Ducky