D.C. circuits

Draw standard symbols and connect meters correctly

Voltmeter connection in parallel

Circuit with ammeter in series and voltmeter connected in parallel across a resistor to measure potential difference.

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Voltmeter measurement topology

Part	Connection
Ammeter	In series in the main loop
Voltmeter	In parallel across the resistor
Resistor	In the conducting loop with the cell and ammeter

Voltmeter measurement topology

Part	Connection
Ammeter	In series in the main loop
Voltmeter	In parallel across the resistor
Resistor	In the conducting loop with the cell and ammeter

Be able to draw a cell or battery, switch, lamp, LED, fixed resistor, variable resistor, fuse, ammeter and voltmeter. Use straight connecting lines and a junction dot only where wires are electrically joined.

Follow nodes, not the shape on the page

A node is a set of points joined by an uninterrupted ideal wire. Bending or stretching that wire does not change the connections. Crossing lines are not enough to establish a connection: read the junction dot or bridge convention. A switch gap breaks the conducting path, while a resistor connects two different nodes.

To place a meter, identify the quantity first. Current describes charge passing through a path; potential difference compares two points. Trace each meter terminal back to the circuit before deciding whether its placement measures the intended quantity.

In series, current is the same and potential differences add

There is one path, so the current is the same at every point:I_source = I₁ = I₂. Energy conservation givesV_source = V₁ + V₂ + ….

In the left-hand diagram, trace the single loop through both resistors. Charge cannot steadily enter the first resistor faster than it leaves the second without accumulating somewhere. The current is therefore common. Each coulomb transfers energy in both resistors, so their energy-per-charge values add to the source value.

Unequal resistances need not share the p.d. equally

Let the two series resistors be 2 Ω and 4 Ω across 12 V. Their total is 6 Ω, giving 2 A throughout. Their p.d.s are 2 × 2 = 4 V and 2 × 4 = 8 V. Check the whole loop: 4 + 8 = 12 V. Equal current does not mean equal p.d.

Opening a switch is different from adding resistance

An open switch breaks the only loop, so there is no steady current anywhere in it. Adding a finite resistance keeps a closed path but increases total resistance, reducing the current for the same ideal supply. Zero current does not require every p.d. to be zero: the open switch can have the supply p.d. across its gap.

In parallel, potential difference is equal and branch currents add

Each branch connects across the same two points, soV_source = V₁ = V₂. Charge conservation at a junction givesI_source = I₁ + I₂ + ….

Return to the right-hand diagram above. Trace the upper wire rail as one node and the lower rail as another. Both resistors have one end on each rail, so each spans the same two electrical potentials. The result depends on their endpoints, not on whether the resistor symbols look parallel on the page.

Split the charge flow and recombine it

With a 12 V ideal supply, branches of 6 Ω and 3 Ω carry 2 A and 4 A. The supply carries 6 A before the split and after the branches rejoin. Check that 2 + 4 = 6 A; do not divide the 12 V between the branches.

Effective resistance: replace a group without changing its terminal behaviour

The equivalent resistance must draw the same total current for the same p.d. across the group's two terminals. First identify those terminals and trace the branches between them. A formula cannot repair a mistaken connection map.

Add series resistances directly

R_total = R₁ + R₂ + …

For 4 Ω and 8 Ω in series, R_total = 12 Ω.

Add reciprocal resistances in parallel

1 / R_total = 1 / R₁ + 1 / R₂ + …

For 6 Ω and 3 Ω in parallel, 1/R = 1/6 + 1/3 = 1/2, soR_total = 2 Ω. The answer is below the smallest branch resistance, as it must be.

The additional conducting path lets more total charge pass each second at the same p.d. Therefore V/I for the whole pair is smaller than for either branch alone. If your answer exceeds the smallest positive branch resistance, check the connection map and whether you took the final reciprocal.

Reduce the network before applying R = V / I

Start at B and follow both paths to C: each contains a 6 Ω resistor. Those resistors share both endpoints, so their equivalent resistance between B and C is 3 Ω. The 4 Ω resistor connects A to B; it is not a third branch between B and C, even though R₀ and R₁ lie on the same horizontal line.

After replacing the B–C pair, the source drives 4 Ω + 3 Ω = 7 Ω. The source current is I = V/R = 14/7 = 2.0 A. R₀ has a p.d. of 2 × 4 = 8 V, leaving 14 − 8 = 6 V between B and C. Each 6 Ω branch therefore carries 1 A; 1 + 1 = 2 A checks the junction.

1. Identify series and parallel groups.
2. Calculate effective resistance from the innermost group outward.
3. Use a p.d., current and resistance that describe the same part of the circuit.
4. Apply current and p.d. rules to recover branch values.

Common mistakes

Check your understanding

Worked example: solve a mixed circuit from the inside out

Trace B to C again. The endpoints are unchanged, but the branch resistances now differ. R₀ is 4 Ω, R₁ is 6 Ω and R₂ is 3 Ω, with an 18 V ideal source.

1. The parallel pair has resistance 2 Ω because 1/R = 1/6 + 1/3 = 1/2.
2. Total resistance = 4 + 2 = 6 Ω.
3. Supply current = 18/6 = 3.0 A. This current passes through the 4 Ω series resistor.
4. P.d. across the 4 Ω resistor = 3.0 × 4 = 12 V.
5. The parallel pair therefore has 18 − 12 = 6.0 V across it.
6. Branch currents are 6.0/6 = 1.0 A and 6.0/3 = 2.0 A; they add to the 3.0 A supply current.

Guided practice

Practise this independently

1. Draw a circuit containing a battery, a switch that controls the whole circuit, two resistors in parallel, an ammeter measuring the supply current and a voltmeter measuring the p.d. across one resistor.
2. Two resistors, 2.0 Ω and 4.0 Ω, are connected in series across 12 V. Find the effective resistance, circuit current and p.d. across each resistor.
3. Two resistors, 6.0 Ω and 3.0 Ω, are connected in parallel across 12 V. Find the effective resistance, each branch current and the supply current.
4. A 5.0 Ω resistor is in series with a parallel pair of 10 Ω and 15 Ω resistors across a 22 V supply. Find the effective resistance, supply current, p.d. across each part and both branch currents.

Try this next

Change one resistance in question 3 and solve it again. Before calculating, predict whether the supply current should rise or fall and check that your final answer agrees.

Practise this topic

The topic check covers circuit symbols and series rules, parallel circuits, and resistance in whole circuits. Use the feedback to return to the matching explanation before trying a fresh question.