Why does adding a parallel resistor reduce the total resistance?

It provides an extra path for charge. With more routes available, the same p.d. drives a larger total current, which means a lower overall resistance.

Why is the current the same everywhere in a series circuit?

Because charge is conserved and there are no junctions. Whatever passes one point must pass every other point in the loop.

How do I approach a circuit with both series and parallel parts?

Reduce it in stages. Combine each parallel group into a single equivalent resistance, then add the series parts, working outwards until one resistance remains.

Do parallel branches always carry equal current?

No, only if their resistances are equal. They share the same p.d., so the current in each is V/R for that branch.

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34–36ElectricityCore

Series and Parallel Circuits

However complicated a circuit looks, two conservation principles govern it: charge cannot accumulate at a junction, and energy must balance around any complete loop. The familiar series and parallel resistor formulas are consequences of those two statements rather than separate rules.

35 min Topic 3 · Electric Circuits

What you'll be able to do

State and apply conservation of charge at a junction
State and apply conservation of energy around a loop
Derive and use the series and parallel resistance formulas
Analyse circuits combining series and parallel sections

Charge conservation at a junction

Charge cannot be created, destroyed, or piled up at a junction, so the total current flowing in equals the total current flowing out. Whatever arrives must leave by some route.

This is why the current is the same at every point in an unbranched loop — there is nowhere else for it to go. A common misconception is that current is "used up" by a lamp; in fact the same current leaves as enters, and what is transferred is energy.

At a branch, the current divides between the routes, with more taking the lower-resistance path.

\Sigma I_{\text{in}} = \Sigma I_{\text{out}}

Conservation of charge at a junction.

Example 1Currents of

0.8\ \mathrm{A}

and

1.5\ \mathrm{A}

flow into a junction, and two branches leave carrying

0.9\ \mathrm{A}

and

I

. Find

I

1Total in:

0.8 + 1.5 = 2.3

2Total out must match:

0.9 + I = 2.3

Answer

I = 1.4\ \mathrm{A}

Tip — If a question describes current being "lost" in a component, that is a distractor. Charge is conserved; only energy is transferred.

Energy conservation around a loop

Each coulomb gains energy at the source and gives it up in the components. Returning to its starting point, its net energy change must be zero — so around any complete loop the e.m.f. equals the sum of the potential differences.

That immediately explains why potential differences in series add up to the supply voltage: each component takes its share of the energy each coulomb carries.

In parallel branches, each branch connects the same two points, so each has the $\textbf{same}$ potential difference across it — the energy given up per coulomb is the same whichever route is taken.

\Sigma \varepsilon = \Sigma V

Energy conservation around a closed loop.

V_{\text{series}} = V_{1} + V_{2} + \ldots

P.d.s add in series.

V_{\text{parallel}} = V_{1} = V_{2}

Same p.d. across parallel branches.

Example 1A 9.0 V supply drives a series loop with three resistors. Two have p.d.s of 2.5 V and 3.5 V. Find the third.

1Energy conservation: the three p.d.s must sum to the e.m.f.

2.5 + 3.5 + V_{3} = 9.0

Answer

V_{3} = 3.0\ \mathrm{V}

Resistors in series

In series the same current $I$ passes through each resistor, and the p.d.s add. So $V_{\text{total}} = IR_{1} + IR_{2} + \ldots$ , and dividing by $I$ gives the total resistance as the simple sum.

Adding a resistor in series always increases the total resistance, because the charge must pass through additional opposition.

A useful check: the total series resistance is always $\textbf{larger}$ than the largest individual resistor.

R_{\text{total}} = R_{1} + R_{2} + \ldots

Series resistances add.

Example 1Resistors of

12\ \Omega

18\ \Omega

and

20\ \Omega

are in series across a 12 V supply. Find the current and the p.d. across the

18\ \Omega

resistor.

1Total resistance:

12 + 18 + 20 = 50\ \Omega

2Current:

I = \dfrac{12}{50} = 0.24

A, the same throughout.

3P.d. across the

18\ \Omega

V = IR = 0.24 \times 18

Answer

I = 0.24\ \mathrm{A}

;

V = 4.32\ \mathrm{V}

Tip — In series, the largest resistance takes the largest share of the p.d. — the resistances and the p.d.s are in the same ratio.

Resistors in parallel

In parallel each branch has the same p.d. $V$ , and the currents add by charge conservation: $I_{\text{total}} = \dfrac{V}{R_{1}} + \dfrac{V}{R_{2}} + \ldots$ . Dividing by $V$ gives the reciprocal formula.

Adding a resistor in parallel always $\textbf{decreases}$ the total resistance, because it provides an additional route for charge — more paths mean easier flow overall.

The check here is the opposite of the series one: the total parallel resistance is always smaller than the smallest individual resistor. An answer larger than any branch means the reciprocal was not inverted at the end.

\dfrac{1}{R_{\text{total}}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \ldots

Parallel resistances combine reciprocally.

R_{\text{total}} = \dfrac{R_{1}R_{2}}{R_{1}+R_{2}}

Shortcut for exactly two resistors.

Example 1Find the total resistance of

6.0\ \Omega

and

3.0\ \Omega

in parallel.

1Using the two-resistor shortcut:

\dfrac{6 \times 3}{6 + 3} = \dfrac{18}{9}

Answer

2.0\ \Omega

— smaller than either resistor, as expected.

Example 2A

10\ \Omega

and a

15\ \Omega

resistor are in parallel across 6.0 V. Find the current in each branch and the total.

1Each branch has the full 6.0 V across it.

I_{1} = \dfrac{6.0}{10} = 0.60

I_{2} = \dfrac{6.0}{15} = 0.40

3Total:

0.60 + 0.40

Answer

0.60\ \mathrm{A}

0.40\ \mathrm{A}

, total

1.0\ \mathrm{A}

Equation recap

\Sigma I_{\text{in}} = \Sigma I_{\text{out}}

Charge conservation.

\Sigma \varepsilon = \Sigma V

Energy conservation round a loop.

R_{s} = R_{1}+R_{2}

Series.

\tfrac{1}{R_{p}} = \tfrac{1}{R_{1}}+\tfrac{1}{R_{2}}

Parallel.

R_{p} = \tfrac{R_{1}R_{2}}{R_{1}+R_{2}}

Two-resistor shortcut.

Common mistakes to avoid

Adding resistances in parallel as if in series.

Use the reciprocal sum — parallel resistance is always less than the smallest branch.

Leaving the answer as

1/R

after a parallel calculation.

Invert to get

R

. An answer of, say,

0.25

when the branches are tens of ohms is unfinished.

Saying current is used up by a lamp.

Charge is conserved; the same current leaves as enters. Energy is what is transferred.

Assuming parallel branches carry equal currents.

They share the same p.d., but the current in each depends on its resistance.

Key takeaways

Charge conservation gives the junction rule; energy conservation gives the loop rule.
Series: same current, p.d.s add, resistances add.
Parallel: same p.d., currents add, resistances combine reciprocally.
Series total exceeds the largest resistor; parallel total is below the smallest.
Reduce mixed circuits step by step rather than seeking a single formula.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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Related lessons

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