Edexcel IAL Physics

2.7 Series and parallel circuits and I–V characteristics

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What this page covers

Overview

Why the current into a junction equals the current out of it (conservation of charge, Kirchhoff's first law)

Why the p.d.s around a closed loop add up to the source p.d. (conservation of energy, Kirchhoff's second law)

Deriving and using R = R₁ + R₂ + R₃ and 1/R = 1/R₁ + 1/R₂ + 1/R₃

A reliable method for currents, p.d.s and resistances in combined series–parallel networks

Sketching, recognising and interpreting the I–V graphs of an ohmic conductor, a filament lamp, an NTC thermistor and a semiconductor diode

Finding resistance from an I–V graph correctly, and solving a circuit containing a non-ohmic component

The circuit and method for measuring an I–V characteristic

In one minute

Key points

1. Two conservation laws do almost all the work in circuit questions: charge is conserved, so current divides at junctions without being used up, and energy is conserved, so the p.d.s around any loop add up to what the source provides.

2. In a series loop the current is the same everywhere; parallel branches all have the same p.d.

3. Series: R = R₁ + R₂ + R₃, larger than any single resistor. Parallel: 1/R = 1/R₁ + 1/R₂ + 1/R₃, smaller than the smallest resistor.

4. I–V graphs: ohmic conductor, a straight line through the origin; filament lamp, gradient falls as it heats; NTC thermistor, gradient rises as it heats; diode, almost no current in reverse or below about 0.6 V, then a steep rise.

5. Resistance at any point is V/I using that point's coordinates, not the gradient.

How is current shared at a junction?

Section 1 of 12

A junction is any point where three or more wires meet. Know the rule and the reason for it: "explain" questions ask for the reason.

Kirchhoff's first law: the sum of the currents into a junction equals the sum of the currents out of it. It is a consequence of conservation of charge.

Current is the rate of flow of charge, I = ΔQ/Δt. Charge cannot be created or destroyed, and in a steady circuit it does not pile up at a junction. So in every second, the charge arriving at a junction must equal the charge leaving it, and dividing both by the time gives exactly the statement about currents. In a series loop, with no junctions, the current is therefore the same at every point. A component transfers energy from the charge; it does not use up charge, so it does not use up current.

How is current shared at a junction? (continued)

Section 1 of 12

At junction J the 0.50 A from the cell splits into 0.30 A through R₁ and 0.20 A towards R₂. At K the branch currents recombine to give 0.50 A again: nothing is lost, it is only shared.

The smaller current flows through the branch with the larger resistance: both branches have the same p.d., so I = V/R gives more current to the smaller resistance.

How is current shared at a junction? (continued)

Section 1 of 12 · Worked example 1 · Easy · Calculate

Label each junction and draw an arrow on every current before you calculate; the unknowns then become simple equations.

WORKED EXAMPLE · QUESTION

A current of 1.50 A enters junction J on a shop's lighting track. From J, branch X carries 0.62 A and branch Y carries 0.35 A; branch Z carries the rest. Further along, branches Y and Z rejoin at junction K. Calculate (a) the current in branch Z, (b) the current leaving K, and (c) the charge that flows through branch Z in 2.0 minutes.

SOLUTION

(a) Current in = current out at J: 1.50 = 0.62 + 0.35 + I_Z, so I_Z = 0.53 A.

(b) Current in = current out at K:

How is current shared at a junction? (continued)

Section 1 of 12 · Worked example 1 · Easy · Calculate (continued)

I_K = 0.35 + 0.53, so I_K = 0.88 A.

(c) Q = It with the time in seconds: Q = 0.53 ×120 = 63.6 C, so Q ≈64 C.

Check: X carries 0.62 A and 0.62 + 0.88 = 1.50 A, so all the current that left J is accounted for.

How is current shared at a junction? (continued)

Section 1 of 12

Common error

Saying the current is "used up" by a lamp or resistor, so it is smaller after the component. In a series loop the current is the same on both sides of every component. What decreases across a component is the potential, because the charge has transferred energy.

How is potential difference shared around a circuit?

Section 2 of 12

Kirchhoff's second law: around any closed loop, the sum of the e.m.f.s equals the sum of the p.d.s. It is a consequence of conservation of energy.

P.d. is energy transferred per unit charge, V = W/Q. A coulomb gains energy in the source, transfers it in the components, and returns to its start with the energy it began with. Energy is conserved, so the energy transferred in the components must equal the energy supplied, and dividing by the charge turns energies into p.d.s. (With negligible internal resistance, e.m.f. = terminal p.d.; see page 2.9.)

Series: the p.d.s across the components add up to the supply p.d. Because the current is the same in each component, V = IR means the p.d. across each resistor is proportional to its resistance.

Parallel: every branch connects the same two points, so every branch has the same p.d. A coulomb passes through only one branch and transfers the same energy whichever branch it takes.

How is potential difference shared around a circuit? (continued)

Section 2 of 12

The series chain of worked example 2. One current flows through all three resistors; the voltmeter is connected across the 200 Ω and 150 Ω resistors together.

A graph of potential against position round the circuit makes the energy picture visible. Connecting wires have negligible resistance, so the potential along a wire is constant. Each resistor causes a drop, and the drops together use up the whole of the source p.d.

How is potential difference shared around a circuit? (continued)

Section 2 of 12

Potential around the series circuit, starting at the positive terminal. Flat sections are wires; each resistor takes a share of the 9.0 V in proportion to its resistance, and the three drops add up to 9.0 V.

How is potential difference shared around a circuit? (continued)

Section 2 of 12 · Worked example 2 · Easy · Calculate

WORKED EXAMPLE · QUESTION

A 9.0 V battery of negligible internal resistance is connected in series with resistors of 100 Ω, 200 Ω and 150 Ω. Calculate (a) the current, (b) the p.d. across each resistor, and (c) the reading of an ideal voltmeter connected across the 200 Ω and 150 Ω resistors together.

SOLUTION

(a) Series total: R = 100 + 200 + 150 = 450 Ω. Then I = V/R = 9.0/450, so I = 0.020 A (20 mA).

(b) V = IR with the same current in each: 0.020 ×100 = 2.0 V, 0.020 ×200 = 4.0 V, 0.020 ×150 = 3.0 V, so 2.0 V, 4.0 V and 3.0 V.

Energy check: 2.0 + 4.0 + 3.0 = 9.0 V, the supply p.d.

(c) The voltmeter spans two resistors, so add their p.d.s: 4.0 + 3.0, giving 7.0 V.

How is potential difference shared around a circuit? (continued)

Section 2 of 12

Exam tip

In a series circuit you can share the supply p.d. in the ratio of the resistances without finding the current: the 200 Ω resistor has (200/450) of 9.0 V, which is 4.0 V. Use the current method when the question asks for the current anyway.

How do you derive the equations for resistors in series and in parallel?

Section 3 of 12

Series: one current, p.d.s add. Parallel: one p.d., currents add. Every resistor equation on this page comes from these two pictures.

Both results come from combining the two conservation laws with V = IR. Learn the derivations as three short steps each; the final equations are in the formulae list, the derivations are not.

Resistors in series

1. The same current I passes through each resistor (conservation of charge).

2. The p.d.s add: V = V₁ + V₂ + V₃ (conservation of energy).

3. Write each p.d. as IR: IR = IR₁ + IR₂ + IR₃, then divide by I.

How do you derive the equations for resistors in series and in parallel? (continued)

Section 3 of 12

R = R₁ + R₂ + R₃

The total is always larger than the largest single resistor: every coulomb has to pass through all of them.

Resistors in parallel

1. The same p.d. V is across each resistor (conservation of energy: each branch joins the same two points).

2. The currents add: I = I₁ + I₂ + I₃ (conservation of charge at the junction).

3. Write each current as V/R: V/R = V/R₁ + V/R₂ + V/R₃, then divide by V.

1/R = 1/R₁ + 1/R₂ + 1/R₃

The total is always smaller than the smallest single resistor.

How do you derive the equations for resistors in series and in parallel? (continued)

Section 3 of 12

Each extra branch is an extra path for charge, so for the same p.d. more current flows, and R = V/I falls.

Useful shortcut	Result	Note
Two resistors in parallel	R = R₁R₂/(R₁ + R₂) ("product over sum")	Not in the formulae list: derive it from the parallel equation. Works for two resistors only.
n identical resistors R₀ in series	nR₀	From the series equation
n identical resistors R₀ in parallel	R₀/n	From the parallel equation
Branch currents in parallel	smaller R carries larger I	Same p.d. across every branch

How do you derive the equations for resistors in series and in parallel? (continued)

Section 3 of 12 · Worked example 3 · Medium · Calculate

Every branch has the full 6.0 V across it, so each branch current is simply 6.0 V divided by that branch's resistance.

WORKED EXAMPLE · QUESTION

Resistors of 22 Ω, 47 Ω and 68 Ω are connected in parallel across a 6.0 V supply of negligible internal resistance. Calculate (a) the combined resistance and (b) the current in each resistor and the supply current.

SOLUTION

(a) Add the reciprocals, then invert: 1/R = (1/22) + (1/47) + (1/68) = 0.0814 Ω⁻¹, so R = 12 Ω (12.3 Ω).

(b) 6.0/22 = 0.273 A, 6.0/47 = 0.128 A, 6.0/68 = 0.0882 A. The supply current is the sum of the branch currents: I = 0.489 A.

How do you derive the equations for resistors in series and in parallel? (continued)

Section 3 of 12 · Worked example 3 · Medium · Calculate (continued)

Check: 6.0 / 12.28 = 0.489 A. And 12 Ω is less than the smallest resistor, 22 Ω, a quick sense check on any parallel answer.

How do you derive the equations for resistors in series and in parallel? (continued)

Section 3 of 12

Common error

Stopping after adding the reciprocals. 1/22 + 1/47 + 1/68 = 0.0814 is 1/R, in Ω⁻¹. Giving "0.081 Ω" as the resistance is wrong by a factor of more than 100, so always invert as the last step.

How do you analyse a combined series–parallel network?

Section 4 of 12

Simplify inwards to a single resistance, find the supply current, then work back outwards. Use the same four steps every time:

1. Replace each parallel group with its single equivalent resistance.

2. Add the series resistances to get the total; find the supply current, I = V/Rₜₒₜₐₗ.

3. Work back out: the p.d. across each series part is I times its resistance, and these p.d.s add up to the supply p.d.

4. Each branch of a parallel group has the group's p.d.; branch current = that p.d. ÷ branch resistance. The branch currents add up to the current into the group.

Changing any one resistor changes every current and p.d. in the network, because the total resistance, and therefore the supply current, changes.

How do you analyse a combined series–parallel network? (continued)

Section 4 of 12

The network of worked example 4. The parallel pair is equivalent to 12 Ω, so the total is 22 Ω. The 2.73 V across the 10 Ω resistor and the 3.27 V across the pair add up to 6.0 V, and 0.164 A + 0.109 A = 0.273 A.

How do you analyse a combined series–parallel network? (continued)

Section 4 of 12 · Worked example 4 · Medium · Calculate

WORKED EXAMPLE · QUESTION

A 6.0 V supply of negligible internal resistance is connected to a 10 Ω resistor in series with a parallel pair of resistors, 20 Ω and 30 Ω. Calculate (a) the total resistance, (b) the supply current, (c) the p.d. across the parallel pair and (d) the current in the 30 Ω resistor.

SOLUTION

(a) Parallel pair first: 1/Rₚ = (1/20) + (1/30) = (5/60), so Rₚ = 12 Ω. Then R = 10 + 12 = 22 Ω.

(b) I = 6.0/22 = 0.273 A.

(c) P.d. across the series resistor: V₁₀ = 0.2727 ×10 = 2.73 V. The rest of the 6.0 V is across the pair: Vₚ = 6.0 - 2.73 = 3.27 V.

(d) I₃₀ = 3.273/30 = 0.109 A.

Check: I₂₀ = 3.273/20 = 0.164 A, and 0.164 + 0.109 = 0.273 A.

How do you analyse a combined series–parallel network? (continued)

Section 4 of 12

A follow-up thought shows how the whole network responds. If the 30 Ω resistor were removed, the total would rise to 10 + 20 = 30 Ω, the supply current would fall to 6.0/30 = 0.20 A, and the p.d. across the 20 Ω resistor would rise to 0.20 ×20 = 4.0 V. Less p.d. is "lost" across the 10 Ω resistor because less current flows through it.

Exam tip

Carry at least one more significant figure than the answer needs through a multi-step network calculation (here 0.2727 A, not 0.27 A). Rounding early can move the final answer in its last figure.

What do the I–V graphs of common components look like?

Section 5 of 12

An I–V characteristic is a graph of the current I through a component against the p.d. V across it. It includes negative values, obtained by reversing the connections, because some components behave differently in the two directions. A component is ohmic if I ∝V: its graph is a straight line through the origin. For a metal this needs constant temperature (Ohm's law, page 2.6).

On I–V axes a steeper line from the origin means a smaller resistance. Why the lamp and thermistor change resistance with temperature is explained on page 2.10.