Resistors in Series and Parallel Combinations

A single resistor obeys $R = \dfrac{V}{I}$. Real circuits contain many, wired in two basic patterns — series and parallel — and every more complicated network is built from these two by nesting. This chapter derives the combination rules from two conservation principles (charge and energy), applies them to ladder networks, and analyses the potential divider, the single most useful circuit idea for sensing and control in the rest of the course.

The method throughout is the same: replace a group of resistors by one equivalent resistor, redraw the simpler circuit, and repeat until a single resistor remains across the supply. Then work backwards, restoring the detail one stage at a time to find the current in, and p.d. across, every component.

Reading note. The two rules are easy to state and easy to misremember. The reliable check: adding a resistor in series always increases total resistance; adding one in parallel always decreases it, to below the smallest resistor present. Any answer that breaks this is wrong.

1. The Two Governing Principles

Every combination rule in this chapter follows from two facts established in Chapter 3.

Charge conservation (Kirchhoff's current law). Charge does not pile up anywhere, so the current into any junction equals the current out of it. In a series path with no junctions, the current is the same everywhere.

Energy conservation (Kirchhoff's voltage law). Each coulomb returns to the supply with the energy it left with, so around any closed loop the sum of the p.d.s across the components equals the emf driving that loop. Across any two components connected between the same two points, the p.d. is the same.

These are stated fully in Chapter 5; here they are used only in the two simple forms above.

2. Resistors in Series

Resistors are in series when they are joined end to end so that the same current passes through each in turn, with no junction between them.

• Current: the same through every resistor, $I$.
• P.d.: the supply p.d. is shared, $V = V_1 + V_2 + V_3 + \dots$ (energy conservation).

Since $V_k = IR_k$ and $V = IR_{\text{series}}$, substituting into $V = V_1 + V_2 + \dots$ and cancelling the common $I$:

The equivalent resistance of a series string is the sum of the individual resistances, and is therefore always larger than the largest one. The p.d. divides in direct proportion to resistance: the biggest resistor takes the biggest share of the voltage.

3. Resistors in Parallel

Resistors are in parallel when both ends of each are connected to the same pair of points, so the same p.d. is across each and the current has a choice of paths.

• P.d.: the same across every resistor, $V$.
• Current: the supply current splits, $I = I_1 + I_2 + I_3 + \dots$ (charge conservation).

Since $I_k = \dfrac{V}{R_k}$ and $I = \dfrac{V}{R_{\text{parallel}}}$, substituting into $I = I_1 + I_2 + \dots$ and cancelling the common $V$:

The reciprocal of the equivalent resistance is the sum of the reciprocals. Consequences worth memorising:

• The parallel combination is always smaller than the smallest resistor in it — adding another path can only make it easier for current to flow.
• For two resistors: $R_{\text{parallel}} = \dfrac{R_1 R_2}{R_1 + R_2}$ (the "product over sum" shortcut, valid for two only).
• For $N$ equal resistors $R$: $R_{\text{parallel}} = \dfrac{R}{N}$.
• The current divides in inverse proportion to resistance: the smallest resistor carries the largest current.

4. Worked Example: A Nested Network

A $12\ \text{V}$ supply of negligible internal resistance is connected to the following: a $4.0\ \Omega$ resistor in series with a parallel pair of $6.0\ \Omega$ and $12\ \Omega$.

Step 1 — combine the parallel pair.

Step 2 — add the series resistor.

Step 3 — total current from the supply.

Step 4 — work backwards. This $1.5\ \text{A}$ flows through the $4.0\ \Omega$ series resistor, dropping $V_{\text{series}} = (1.5)(4.0) = 6.0\ \text{V}$ across it. The remaining $12 - 6.0 = 6.0\ \text{V}$ is across the parallel pair. So:

These sum to $1.5\ \text{A}$, checking against the supply current.

5. The Potential Divider

Two resistors in series across a supply form a potential divider: the output is the p.d. tapped off across one of them.

With $R_1$ and $R_2$ in series across a supply p.d. $V_{\text{in}}$, the current is $I = \dfrac{V_{\text{in}}}{R_1 + R_2}$, so the p.d. across $R_2$ is

The output is the input scaled by the fraction of the total resistance that $R_2$ represents. Making one resistor a sensor — a thermistor or a light-dependent resistor (LDR) — turns the divider into a transducer: as the temperature or light level changes, that resistance changes, and $V_{\text{out}}$ swings in response, ready to trigger a downstream circuit.

Worked example. A $9.0\ \text{V}$ supply is connected across a $2.0\ \text{k}\Omega$ fixed resistor in series with a thermistor. At room temperature the thermistor is $4.0\ \text{k}\Omega$; when warmed it falls to $1.0\ \text{k}\Omega$. Find $V_{\text{out}}$ across the fixed resistor at both temperatures.

Room temperature:

Warmed:

Warming the thermistor doubles the output — the mechanism behind a temperature-activated fan or alarm.

6. Loading a Potential Divider

If a load (say a voltmeter of resistance $R_V$, or the next stage of a circuit) is connected across the output resistor $R_2$, it is now in parallel with $R_2$. The effective lower resistor becomes $\dfrac{R_2 R_V}{R_2 + R_V}$, which is smaller than $R_2$, so $V_{\text{out}}$ drops below its unloaded value. The divider is only accurate when the load resistance is much larger than $R_2$. This is why a good voltmeter has a very high resistance — so that measuring the output barely disturbs it.

Common Exam Pitfalls

• Adding parallel resistances directly. $\dfrac{1}{R_p}$ is the sum of reciprocals; the final step of taking the reciprocal is the one most often forgotten, leaving the answer as $\dfrac{1}{R_p}$ instead of $R_p$.
• Using "product over sum" for three or more resistors. That shortcut is valid for exactly two. For three, either use the full reciprocal sum or combine two at a time.
• Assuming equal p.d. across series resistors or equal current through parallel ones. It is the reverse: series share current, parallel share p.d.
• Forgetting to check the total against the components. Series total must exceed every part; parallel total must be less than every part.
• Treating a loaded potential divider as unloaded. A load across the output resistor is in parallel with it and pulls the output voltage down.
• Mismatched units. Keep resistances in the same unit throughout ($\text{k}\Omega$ with $\text{mA}$ and $\text{V}$ is a consistent set; mixing $\Omega$ and $\text{k}\Omega$ is a factor-of-1000 error).

Summary

• Series: same current through each; p.d.s add; $R_{\text{series}} = R_1 + R_2 + \dots$, always larger than the largest resistor. P.d. divides in direct proportion to resistance.
• Parallel: same p.d. across each; currents add; $\dfrac{1}{R_{\text{parallel}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$, always smaller than the smallest resistor. Current divides in inverse proportion to resistance.
• Two resistors in parallel: $R_p = \dfrac{R_1 R_2}{R_1 + R_2}$. $N$ equal in parallel: $\dfrac{R}{N}$.
• Method: reduce groups to equivalents, redraw, solve the simple circuit, then work backwards restoring detail; check currents sum at junctions.
• Potential divider: $V_{\text{out}} = V_{\text{in}}\dfrac{R_2}{R_1 + R_2}$. Replacing one resistor with a thermistor or LDR makes a sensing circuit; a low-resistance load across the output reduces $V_{\text{out}}$.