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Current and potential difference in series and parallel circuits

Key idea: Series components carry the same current, and their potential differences add. Parallel branches have the same potential difference, and their currents add to the supply current.

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Unscrew one bulb from an old string of fairy lights and they all go out; switch off one lamp at home and the others stay on. The first is a series circuit and the second is parallel. Four rules, two for each kind, give every unknown current and potential difference, using p.d. as the energy transferred per unit charge from E.m.f., potential difference and resistance.

Series and parallel circuit rulesTwo circuit diagrams. The series circuit has two resistors in one loop labelled with the same current I and potential differences V1 and V2 that add to the supply. The parallel circuit has two resistor branches, both across supply V, with currents I1 and I2 that add to the source current I.Series circuitone pathR₁, p.d. V₁R₂, p.d. V₂II is the same; V₁ + V₂ = VParallel circuitseparate branches between the same two pointsR₁R₂I₁I₂V₁ = V₂ = V; I = I₁ + I₂
Series and parallel circuit rulesTwo circuit diagrams. The series circuit has two resistors in one loop labelled with the same current I and potential differences V1 and V2 that add to the supply. The parallel circuit has two resistor branches, both across supply V, with currents I1 and I2 that add to the source current I.Series circuitone pathR₁, p.d. V₁R₂, p.d. V₂II is the same; V₁ + V₂ = VParallel circuitseparate branches between the same two pointsR₁R₂I₁I₂V₁ = V₂ = V; I = I₁ + I₂
Series has one path: current is unchanged and component p.d.s add. Parallel branches share the same two endpoints: branch p.d.s are equal and branch currents add at each junction.

In a series circuit the components are joined one after another in a single loop, so there is only one path for the charge. In a parallel circuit the components are in separate branches, and every branch connects the same two junctions. Assume the source p.d. stays fixed, the wires have negligible resistance and the meters are ideal.

Series: the current is the same at every point

Rule: the current at every point in a series circuit is the same.

I = I₁ = I₂ = I₃

There is only one path, so every charge that passes through the first component must pass through the next one. In steady conditions, charge is not accumulating anywhere in the wire, so the charge passing each point per second, the current, is the same everywhere in the loop. It does not matter where you put the ammeter.

The components do not use up charge. The charge carries energy from the source, and each component transfers some of that energy. The same charge flows back into the source.

Check your understanding 1

A learner claims that 0.30 A enters a series resistor but 0.25 A leaves it for 20 s. Calculate how much charge would accumulate. Can this describe a steady current?

Show the steps Optional working
Unit: A
Unit: C
Steady current possible?
Show answer

The difference is 0.30 − 0.25 = 0.05 C each second, so 0.05 × 20 = 1.0 C would accumulate. Charge cannot keep building up in a steady circuit: the same charge must enter and leave each second. The two current values must therefore be equal; the resistor transfers energy without using up charge.

Series: the p.d.s add up to the p.d. across the whole circuit

Rule: the sum of the potential differences in a series circuit is equal to the potential difference across the whole circuit.

V = V₁ + V₂ + V₃

Each coulomb gains energy in the source and transfers it, part by part, in the components. The energy per coulomb transferred in all the components adds up to the energy per coulomb the source gives, so the p.d.s add up to the source’s p.d.

The rule does not say the p.d. is shared equally. With a current of 2.0 A through a 2.0 Ω resistor and a 4.0 Ω resistor in series, the p.d.s are V = IR = 2.0 × 2.0 = 4.0 V and 2.0 × 4.0 = 8.0 V. They add up to 12 V. The same current through a larger resistance needs a larger p.d.

Worked example 1

Three components in series

Problem

Three components in one series loopA 9.0 V battery drives one series loop containing a lamp, resistor X and resistor Y along the top wire, and an ammeter on the bottom wire that reads 0.50 A. Voltmeter V1 is connected across the lamp and reads 2.5 V. Voltmeter V2 is connected across resistor X and reads 4.0 V. No meter is connected across resistor Y.AVV9.0 VlampXYV₁: 2.5 VV₂: 4.0 V0.50 A
Three components in one series loopA 9.0 V battery drives one series loop containing a lamp, resistor X and resistor Y along the top wire, and an ammeter on the bottom wire that reads 0.50 A. Voltmeter V1 is connected across the lamp and reads 2.5 V. Voltmeter V2 is connected across resistor X and reads 4.0 V. No meter is connected across resistor Y.AVV9.0 VlampXYV₁: 2.5 VV₂: 4.0 V0.50 A
One loop, so one current. The p.d.s across the lamp, X and Y add up to the 9.0 V across the battery.

A 9.0 V battery drives a lamp, resistor X and resistor Y in series. The ammeter reads 0.50 A. Voltmeter V₁ reads 2.5 V across the lamp and V₂ reads 4.0 V across X. Find the p.d. across Y, the current in Y and the resistance of Y.

Show full solution

  1. Choose the rule

    Method

    Check how the components are connected.

    Reason

    The three components are in one loop, so the series rules apply.

    Working

    Answer

    Series circuit

  2. p.d. across Y

    Method

    Subtract the other p.d.s from the supply.

    Reason

    The p.d.s add up to 9.0 V.

    Working

    Answer

    VY = 9.0 - 2.5 - 4.0 = 2.5 V

  3. Current in Y

    Method

    Read the ammeter.

    Reason

    The current is the same everywhere in the loop.

    Working

    Answer

    IY = 0.50 A

  4. Resistance of Y

    Method

    Use the p.d. across Y and the current through Y.

    Reason

    Both readings must belong to the same component.

    Working

    Answer

    R = V/I = 2.5/0.50 = 5.0 Ω

  5. Check

    Working

    Answer

    2.5 + 4.0 + 2.5 = 9.0 V. ✓

Check your understanding 2

Lamps P and Q are in series. P transfers 1.8 J electrically each second; Q transfers 0.90 J each second. Compare their currents and calculate the ratio of P’s p.d. to Q’s.

Current in P compared with Q
Show answer

The current is the same: there is one unbranched path. The same charge passes each lamp each second, but P transfers twice the energy per second, so each coulomb transfers twice the energy in P. P therefore has twice Q’s p.d. The same charge transfers twice as much energy in P.

Opening a switch in a series circuit

An open switch breaks the only path, so the current stops everywhere in the loop and every lamp goes out. Adding another resistive lamp in series keeps the loop complete but increases total resistance. With the same ideal supply voltage, the current is smaller everywhere. Original lamps whose electrical properties remain unchanged transfer less energy each second and become dimmer.

Parallel: every branch has the same p.d.

Rule: the potential difference across the separate branches of a parallel circuit is the same.

V = V₁ = V₂ = V₃

Trace the wire from one end of each branch: it reaches the same junction. Trace the other end: it reaches the other junction. Every branch spans the same two points, so each branch has the same p.d. The rule depends on where each branch’s ends are connected, not on whether the branches look parallel on the page.

Equal p.d. does not mean equal current. A branch with a smaller resistance carries a larger current at the same p.d.

Parallel: the branch currents add up to the source current

Rule: the sum of the currents in the separate branches of a parallel circuit is equal to the current from the source.

I = I₁ + I₂ + I₃

At a junction the charge splits between the branches; at the next junction it recombines. Charge is not created or lost, so the charge per second going into a junction equals the charge per second coming out.

With a 12 V supply, branches of 6.0 Ω and 3.0 Ω carry 12/6.0 = 2.0 A and 12/3.0 = 4.0 A. The supply current is 2.0 + 4.0 = 6.0 A, before the split and after the branches rejoin. Do not divide the 12 V between the branches.

Worked example 2

Three parallel branches

Problem

Three parallel branchesA 12 V battery on the left. Ammeter A in the main wire reads 1.5 A. The wire then splits at a junction into three branches that rejoin at a second junction before returning to the battery. Each branch contains its own ammeter and resistor: A1 reads 0.40 A in the branch with R1, A2 reads 0.60 A in the branch with R2, and the reading of A3 in the branch with R3 is unknown.AA: 1.5 A12 VAA₁: 0.40 AR₁AA₂: 0.60 AR₂AA₃: ?R₃
Three parallel branchesA 12 V battery on the left. Ammeter A in the main wire reads 1.5 A. The wire then splits at a junction into three branches that rejoin at a second junction before returning to the battery. Each branch contains its own ammeter and resistor: A1 reads 0.40 A in the branch with R1, A2 reads 0.60 A in the branch with R2, and the reading of A3 in the branch with R3 is unknown.AA: 1.5 A12 VAA₁: 0.40 AR₁AA₂: 0.60 AR₂AA₃: ?R₃
Every branch connects the same two junctions. The branch currents add up to the current in the main wire.

A 12 V battery drives three resistors in parallel. Ammeter A in the main wire reads 1.5 A; A₁ reads 0.40 A and A₂ reads 0.60 A. Find the reading on A₃, the p.d. across R₃ and the resistance R₃.

Show full solution

  1. Choose the rule

    Method

    Check how the branches are connected.

    Reason

    The branches connect the same two junctions, so the parallel rules apply.

    Working

    Answer

    Parallel circuit

  2. Current in the third branch

    Method

    Subtract the other branch currents from the main current.

    Reason

    The branch currents add up to 1.5 A.

    Working

    Answer

    I₃ = 1.5 - 0.40 - 0.60 = 0.50 A

  3. p.d. across R₃

    Method

    Use the supply p.d.

    Reason

    Every branch has the battery’s p.d.

    Working

    Answer

    V₃ = 12 V

  4. Resistance

    Method

    Divide the branch p.d. by the branch current.

    Reason

    Both readings belong to R₃.

    Working

    Answer

    R₃ = V/I = 12/0.50 = 24 Ω

  5. Check

    Working

    Answer

    0.40 + 0.60 + 0.50 = 1.5 A. ✓ The branch with the largest current, R₂, has the smallest resistance: 12/0.60 = 20 Ω.

Check your understanding 3

Four identical lamps are connected in parallel across 6.0 V. The supply current is 1.2 A. Find each lamp’s current and p.d.

Unit: A
Unit: V
Show answer

Each lamp has the full 6.0 V. The lamps are identical, so they carry equal currents: 1.2/4 = 0.30 A each.

Opening, removing or adding a branch

Treat the supply as ideal: its p.d. stays the same whatever the circuit draws.

  • Open one branch. That branch now carries no current. The other branches still connect the same two junctions, so they keep the same p.d. and the same current. The supply current falls by the current the open branch used to carry.
  • Add a branch. The new branch also has the full p.d. across it. The existing branches are unchanged. The supply current rises by the new branch’s current.

With 6.0 Ω and 3.0 Ω in parallel across 12 V, opening the 6.0 Ω branch leaves the 3.0 Ω branch carrying 4.0 A, and the supply current falls from 6.0 A to 4.0 A. This is why the lights in a house are connected in parallel: each lamp gets the full mains p.d., and switching one off does not affect the others.

Check your understanding 4

Resistors of 6.0 Ω and 3.0 Ω are in parallel across 12 V. A 12 Ω branch is added; the supply p.d. stays fixed. Find the new supply current and the current in the 3.0 Ω branch.

Show the steps Optional working
Unit: A
Unit: A
Unit: A
Unit: A
Unit: A
Show answer

The new branch carries 12/12 = 1.0 A, so the supply current rises from 6.0 A to 2.0 + 4.0 + 1.0 = 7.0 A. The 3.0 Ω branch still has 12 V across it, so it still carries 4.0 A.

Series and parallel side by side

SeriesParallel
Paths for the chargeOneOne per branch
CurrentThe same everywhere: I = I₁ = I₂Branch currents add: I = I₁ + I₂
p.d.Component p.d.s add: V = V₁ + V₂The same across every branch: V = V₁ = V₂
One component switched off or removedThe current stops everywhereOnly that branch stops; the others are unchanged
One more component addedSmaller current everywhereA new branch with the full p.d.; the supply current rises

Before using a rule, trace the circuit. Components are in series only if nothing branches off between them. Components are in parallel only if both of their ends connect to the same two junctions.

Check your understanding 5 Use endpoints to identify the groups

A 12 V source connects A to C. X connects A to B, Y connects B to C and Z connects A to C; nothing else connects to B. Sketch the circuit on paper. Classify the connections, then find Z’s p.d. and the sum of X’s and Y’s p.d.s.

X and Y
Z is parallel with
Unit: V
Unit: V
Show answer

Compare your sketch: one A-to-C path goes through X, then B, then Y; the other goes through Z. Mark A and C as junctions, with no extra branch at B. X and Y are in series because the same branch current passes through them. Z is parallel with the whole X–Y group, not X alone: X ends at A and B, while Z ends at A and C. Z has 12 V across it; X’s p.d. plus Y’s p.d. equals 12 V. No individual X or Y p.d. follows without further information.

Common mistakes

  • Saying current is used up as it goes round a series circuit. Charge is not used up; the current is the same at every point in the loop.
  • Saying the p.d. is shared equally in series. The share depends on each component’s resistance; use V = IR for each, and the p.d.s add up to the total.
  • Dividing the supply p.d. between parallel branches. Each branch has the full p.d.; it is the current that divides.
  • Saying parallel branches carry equal currents. The branch with less resistance carries more current.
  • Saying opening one parallel branch turns everything off. Only that branch stops; the supply current falls.
  • Calling components drawn side by side parallel. Trace both ends of each component to its junctions.

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