Parallel resistors: deriving 1/R = 1/R₁ + 1/R₂

The main symbols used in this lesson:

Two components are in parallel when they are connected across the same two points, so each one has its own separate path for charge.

Both of Kirchhoff's laws have something to say about this arrangement.

• Kirchhoff's second law gives the p.d.: each branch connects the same two points, so the p.d. across every branch is the same.
• Kirchhoff's first law gives the current: the current splits at the junction, and the branch currents add back up to the total current.

Connect two resistors $R_1$ and $R_2$ in parallel across a source of e.m.f. $V$. Call the total current leaving the source $I$.

1. Same p.d. (second law). Both branches are connected between the same two points, so the p.d. across each is $V$.
2. Branch currents from $V = IR$. Each branch obeys the definition of resistance, so $I_1 = \dfrac{V}{R_1}$ and $I_2 = \dfrac{V}{R_2}$.
3. Currents add (first law). At the junction, $I = I_1 + I_2$.
4. Substitute and match forms. $I = \dfrac{V}{R_1} + \dfrac{V}{R_2} = V\left(\dfrac{1}{R_1} + \dfrac{1}{R_2}\right)$. The pair behaves like one resistor $R$ with $I = \dfrac{V}{R}$, so comparing the two expressions gives $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$.

For three or more resistors in parallel, extend the sum: $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dots$