Resistors in Series and in Parallel

One current through every resistor

Join two resistors end to end, with no junction between them. Charge that leaves the first resistor has nowhere to go except into the second. So in steady flow both carry the same current \(I\). Resistors joined this way are in series.

Each resistor takes a potential difference given by Ohm's law, from the chapter Electric Current and Resistance: \(V_1 = IR_1\) and \(V_2 = IR_2\). Walking from one end of the chain to the other, the potential falls by \(V_1\) and then by \(V_2\). So the whole chain takes \(V = V_1 + V_2\) \(= I(R_1 + R_2)\). A single resistor \(R_1 + R_2\) would draw the same current from the same voltage. That single resistor is the equivalent resistance of the chain.

So \(2\,\Omega\) and \(3\,\Omega\) in series make \(5\,\Omega\). Three \(6\,\Omega\) resistors in series make \(18\,\Omega\). A series chain is always larger than its largest member. Every resistor adds more obstacle to the one path.

The voltage divider

In series, each resistor takes a share of the voltage in proportion to its resistance. The current is \(I = V/R_s\), so \(V_1 = IR_1\) \(= VR_1/R_s\). Two resistors of \(1\,\text{k}\Omega\) and \(2\,\text{k}\Omega\) across \(12\,\text{V}\) take \(4\,\text{V}\) and \(8\,\text{V}\). A pair used to pick off a fixed fraction of a voltage is called a voltage divider. The volume knob of an old radio is a divider whose tapping point slides.

One voltage across every resistor

Now join each resistor between the same two junctions, A and B. A wire of negligible resistance has one potential all along it. So every resistor has one end at \(V_A\) and the other at \(V_B\), and all take the same potential difference \(V\). Resistors joined this way are in parallel.

The current from the battery reaches junction A and divides among the resistors, then joins again at B. Charge is conserved, so \(I = I_1 + I_2\) \(= V/R_1 + V/R_2\). One resistor \(R_p\) that draws the same \(I\) at the same \(V\) has \(1/R_p = I/V\). So \(1/R_p = 1/R_1 + 1/R_2\).

For \(2\,\Omega\) and \(3\,\Omega\) in parallel, \(1/R_p\) \(= 1/2 + 1/3\) \(= 5/6\), so \(R_p = 1.2\,\Omega\). The result is smaller than the smaller resistor. Adding a branch gives the current one more path, so the network lets more current through at the same voltage.

How the current divides

In parallel the same \(V\) drives each branch, so a branch of smaller resistance takes more current. With two branches, \(I_1/I_2 = R_2/R_1\). Resistors of \(4\,\Omega\) and \(12\,\Omega\) sharing \(4\,\text{A}\) take it as \(3\,\text{A}\) and \(1\,\text{A}\). Both then have \(12\,\text{V}\) across them, as they must.

Take \(n\) equal resistors \(R\). In series they make \(nR\); in parallel, \(R/n\). The ratio of the two is \(n^2\). Five equal resistors give a series value \(25\) times their parallel value.

That is the answer to the Diwali string. In series, the bulbs share one path, so one broken filament stops the current everywhere. The wiring of a house is in parallel: every lamp and fan sits across the same \(220\,\text{V}\) mains. A fused bulb removes one branch, and the others keep their full voltage and their own currents. Fig 1.1 shows both.

Reducing a network step by step

Most networks mix the two. Reduce them from the inside out. Find a group that is clearly series or clearly parallel, replace it by one resistor, and redraw. Repeat until one resistor is left. Then work backwards, spreading the current out again.

In Fig 1.2 the \(6\,\Omega\) and \(3\,\Omega\) share both ends, so they are in parallel: \(6 \times 3/9 = 2\,\Omega\). This \(2\,\Omega\) is in series with the \(4\,\Omega\), giving \(6\,\Omega\). The cell drives \(18/6 = 3\,\text{A}\). The \(4\,\Omega\) takes \(12\,\text{V}\), leaving \(6\,\text{V}\) across the pair. So \(1\,\text{A}\) flows through the \(6\,\Omega\) and \(2\,\text{A}\) through the \(3\,\Omega\), which add back to \(3\,\text{A}\).

Two resistors give \(9\,\Omega\) in series and \(2\,\Omega\) in parallel. Then \(R_1 + R_2 = 9\) and \(R_1R_2/(R_1 + R_2) = 2\), so \(R_1R_2 = 18\). Two numbers with sum \(9\) and product \(18\) are \(3\) and \(6\). Similarly, to make \(2\,\Omega\) from a \(3\,\Omega\) resistor, join \(6\,\Omega\) across it, since \(1/2 - 1/3 = 1/6\).

With \(2\), \(4\) and \(6\,\Omega\), all three in series give the largest value, \(12\,\Omega\). All three in parallel give the smallest: \(1/2 + 1/4 + 1/6\) \(= 11/12\), so \(12/11\,\Omega\). Any other arrangement lies between the two. Resistors of \(1\), \(2\) and \(3\,\Omega\) in parallel give \(6/11\,\Omega\).

A wire reshaped is a network in disguise

A uniform wire of resistance \(R\) has \(R = \rho L/A\), from the chapter Electric Current and Resistance. Cut it into three equal pieces and each has \(R/3\). Join the three side by side, and they form three equal resistors in parallel, \((R/3)/3 = R/9\). A \(9\,\Omega\) wire so treated gives \(1\,\Omega\).

Folding is the same thing. A \(20\,\Omega\) wire folded in half is two \(10\,\Omega\) strands in parallel, \(5\,\Omega\). Folded into four, it is four \(5\,\Omega\) strands in parallel, \(1.25\,\Omega\). Cutting or folding into \(n\) divides the resistance by \(n^2\): the path is \(n\) times shorter and \(n\) times wider.

Stretching keeps the volume \(AL\) fixed. Twice the length means half the area, so \(R = \rho L/A\) grows by \(2 \times 2 = 4\). A \(4\,\Omega\) wire drawn to twice its length has \(16\,\Omega\). If instead its radius is halved, its area falls to a quarter and its length grows four times. Then \(R\) grows \(16\) times, to \(64\,\Omega\).

Two wires of one material and one length, with radii \(r\) and \(2r\), have areas in the ratio \(1 : 4\). So their resistances are in the ratio \(4 : 1\), thin to thick. If the thin one has \(4\,\Omega\), the thick one has \(1\,\Omega\), and together in parallel they give \(4 \times 1/5 = 0.8\,\Omega\).