Resistors in Series and Parallel

Problem

Consider three resistors with different resistivity $R_1$, $R_2$ and $R_3$.

Part 1: Determine the equivalent resistance when the three resistors are connected in series.

Part 2: Determine the equivalent resistance when the three resistors are connected in parallel.

Solution

Part 1: Resistors connected in series (same current, different voltage)

Let $V$ represents the voltage of the battery that follows Ohm's rule $V = I{R}_{eq}$. Here ${R}_{eq}$ represents the equivalent resistance of the resistors. When two resistors are connected in series, the current going through each resistor is the same, and the voltage drop across each resistor is proportional to their respective resistances. By Kirchoff's loop rule,

$V - {I}_{1}{R}_{1} - {I}_{2}{R}_{2} - {I}_{3}{R}_{3} = 0$

$V = {I}_{1}{R}_{1} + {I}_{2}{R}_{2} + {I}_{3}{R}_{3}$

$I{R}_{eq} = {I}_{1}{R}_{1} + {I}_{2}{R}_{2} + {I}_{3}{R}_{3}$

Since $I_1 = I_2 = I_3 = I$,

$I{R}_{eq} = I\left({R}_{1} + {R}_{2} + {R}_{3}\right)$.

Therefore,

${R}_{eq} = {R}_{1} + {R}_{2} + {R}_{3}$

In general, given $n$ resistors added in series

${R}_{eq} = {R}_{1} + {R}_{2} + {R}_{3} + ... + {R}_{n}$

Part 2: Resistors connected in parallel (same voltage, different current)

When two resistors are connected in parallel, the voltage going through each resistor is the same, and the current the enters into the parallel branch is distributed across each resistor inversely-proportional to each resistor. By Kirchoff's junction rule,

$I = I_1 + I_2 + I_3$

$\frac{V}{{R}_{eq}} = \frac{{V}_{1}}{{R}_{1}} + \frac{{V}_{2}}{{R}_{2}} + \frac{{V}_{3}}{{R}_{3}}$

Since $V_1 = V_2 = V_3 = V$,

$\frac{V}{{R}_{eq}} = \frac{V}{{R}_{1}} + \frac{V}{{R}_{2}} + \frac{V}{{R}_{3}}$.

Therefore,

$\frac{1}{{R}_{eq}} = \frac{1}{{R}_{1}} + \frac{1}{{R}_{2}} + \frac{1}{{R}_{3}}$.

In general, given $n$ springs added in parallel

$\frac{1}{{R}_{eq}} = \frac{1}{{R}_{1}} + \frac{1}{{R}_{2}} + \frac{1}{{R}_{3}}+ ... + \frac{1}{{R}_{n}}$.