Resistors in series and parallel calculator

• Series: 4 Ω and 6 Ω. $R_T = 4 + 6 = 10$ Ω.
• Parallel, two resistors: 4 Ω and 6 Ω. $\dfrac{1}{R_T} = \dfrac{1}{4} + \dfrac{1}{6} = 0.4167$, so $R_T = 2.4$ Ω.
• Parallel, three resistors: 6 Ω, 6 Ω and 3 Ω. $\dfrac{1}{R_T} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{2}{3}$, so $R_T = 1.5$ Ω.

For exactly two resistors in parallel there's a shortcut:

For 4 Ω and 6 Ω: $24 \div 10 = 2.4$ Ω, the same answer. Two equal resistors in parallel give half the resistance of one.

Use the total resistance with Ohm's law to find the current from the supply.

For a combination, work from the inside out: find the total of each parallel group first, then add it to the series resistors. For a 2 Ω resistor in series with a parallel pair of 4 Ω and 6 Ω: the pair gives 2.4 Ω, so the total is $2 + 2.4 = 4.4$ Ω.

• Forgetting the last step in parallel: $\tfrac{1}{R_T}$ isn't the answer, flip it.
• Using product over sum for three or more resistors.
• Adding parallel resistances as if they were in series.

Practise with our series and parallel circuit questions and read our guide to series and parallel circuits.

• Redraw complicated circuits so each parallel group is clear.
• Show the $\tfrac{1}{R}$ line and the "flip" line separately.
• In parallel, check your total is less than the smallest resistor; in series, more than the largest.
• Adding a resistor in parallel always lowers the total resistance, which is why plugging more devices into a parallel circuit draws more current.