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Topic 4.3 · Unit 4

Electric circuits

Circuit components and how they behave, the rules for current, p.d. and resistance in series and parallel circuits, and how a potential divider shares a p.d.

In this topic

  1. 4.3.1Circuit diagrams and circuit components
  2. 4.3.2Series and parallel circuits
  3. 4.3.3Action and use of circuit components

Key points

4.3.1 Circuit diagrams and circuit components

  • A circuit diagram shows how components are connected. Each component has its own symbol. Draw connecting wires as straight lines with a ruler.
  • The table describes each symbol in words and says how each component behaves in a circuit.
Component Symbol (in words) What it does in a circuit
cell a long thin line and a short thick line; the long line is the + terminal provides an e.m.f.; gives d.c.
battery two or more cell symbols joined in a row two or more cells in series; a larger e.m.f.
power supply two small open circles (the terminals) with a gap between them, each at the end of a connecting line provides a p.d. for the circuit
d.c. power supply the power supply symbol with + above one terminal and − above the other provides a d.c. p.d.
a.c. power supply the power supply symbol with a ~ (wave) sign between the two terminals provides an a.c. p.d.
generator a square box with G inside provides an e.m.f. when it is turned (see 4.5.2)
switch a gap in the line with a short line hinged at one side open: no current; closed: current can flow
fixed resistor a small rectangle its resistance limits the current
variable resistor a rectangle with a diagonal arrow through it its resistance can be changed, to change the current
potential divider a rectangle with an arrow pointing at its side a sliding contact gives an output p.d. that can be changed
heater a rectangle divided into several sections transfers electrical energy into thermal energy
thermistor (NTC) a rectangle with a sloping line through it; the line has a short flat end its resistance decreases as its temperature increases
light-dependent resistor (LDR) a rectangle with two arrows pointing in towards it its resistance decreases as the light on it gets brighter
lamp a circle with a cross inside lights up when a current flows
motor a circle with M inside turns when a current flows
electric bell a dome (a half circle, flat side down) with two straight leads going down from the flat side rings when a current flows
ammeter a circle with A inside measures current; connected in series
voltmeter a circle with V inside measures p.d.; connected in parallel
magnetising coil a row of loops, like a spring becomes a magnet when a current flows
transformer two coils side by side with a straight line (the core) between them changes the size of an a.c. voltage (see 4.5.6)
fuse a rectangle with a line running through it from end to end melts and breaks the circuit if the current is too large (see 4.4)
relay coil a small rectangle with one lead from the top and one from the bottom a small current in its coil switches a second circuit on or off (see 4.5.3)
junction of conductors a dot where two or more lines meet the wires are joined there; current can split or join
earth (ground) a vertical line down to three short horizontal lines, each shorter than the one above a connection to the ground
  • NTC stands for “negative temperature coefficient”. It means the resistance goes down as the temperature goes up. A thermistor can be used as a temperature sensor.
  • An LDR can be used as a light sensor.
  • Diode: its symbol is a triangle pointing at a short line across the wire. Extended
  • A diode lets current flow in one direction only: the direction the triangle points (conventional current). In the other direction its resistance is very high, so almost no current flows. Extended
  • Light-emitting diode (LED): a diode symbol with two small arrows pointing away from it. An LED gives out light when current flows through it in its forward direction. It does not light if it is connected the wrong way round. Extended
  • Extra detail: because a diode lets current through one way only, it can be used to change a.c. into d.c.

4.3.2 Series and parallel circuits

  • In a series circuit, the components are connected one after another in a single loop. There is only one path for the current.
  • In a parallel circuit, the components are on separate branches. Each branch is connected across the supply. The current splits between the branches.
  • Reading a circuit diagram (example): a diagram shows a cell, a switch and two lamps, L1 and L2.
    • The switch is next to the cell, in the main part of the circuit.
    • After the switch, the wire reaches a junction (a dot) and splits into two branches. One branch holds L1. The other holds L2.
    • The two branches join again at a second junction, and the wire goes back to the other terminal of the cell.
    • So each lamp’s branch is connected across the cell. Each lamp has the full p.d. of the cell.
    • The switch is in the main part of the circuit, so it turns both lamps on and off together.
    • If L1 is removed, its branch is broken. The branch with L2 is still a complete path, so L2 stays on. The current from the cell gets smaller, because only one branch now carries current.
    • If the two lamps were in series instead, removing one would break the only path. Both lamps would go out.
  • The current is the same at every point in a series circuit. An ammeter gives the same reading wherever it is placed in the loop.
  • Sources in series: when cells are joined in series, the same way round (+ of one to − of the next), their e.m.f.s add up.
  • If one cell is turned the wrong way round, its e.m.f. is subtracted instead.
  • Resistors in series: the combined resistance is the sum of the separate resistances: R = R1 + R2 + …
  • In a parallel circuit, the current from the source is larger than the current in any one branch.
  • The combined resistance of two resistors in parallel is less than the resistance of either one alone. Adding a second branch gives the current another path, so more current flows from the supply.
  • Lamps in a lighting circuit (for example the lights in a house or a classroom) are connected in parallel. The advantages:
    • each lamp gets the full supply p.d., so each lamp is fully bright
    • each lamp can be switched on and off on its own
    • if one lamp breaks, the others stay on.
  • Junction rule: the sum of the currents entering a junction is equal to the sum of the currents leaving it. Extended
  • Why: charge cannot be made, destroyed or stored at a junction. So every second, the charge that flows in must equal the charge that flows out. Current is charge per second, so current in = current out. Extended
  • In a series circuit, the p.d.s add up: the total p.d. across the components equals the sum of the p.d.s across each one. Extended
  • In a parallel arrangement, the p.d. across the whole arrangement is the same as the p.d. across each branch. Extended
  • Combined resistance of two resistors in parallel: R = R1R2 / (R1 + R2), or 1 / R = 1 / R1 + 1 / R2. Extended

4.3.3 Action and use of circuit components

  • V = IR. So for a constant current, the p.d. across a conductor increases when its resistance increases.
  • In a series circuit, the current is the same in each resistor. So the resistor with the larger resistance has the larger p.d. across it.
  • A potential divider is two resistors in series across a supply. The supply p.d. is shared between them in the ratio of their resistances: R1 / R2 = V1 / V2. Extended
  • A variable potential divider gives an output p.d. that can change. One resistor is variable, or it is a thermistor or an LDR. The output p.d. is taken across one of the two resistors. Extended
  • If the resistance of one resistor increases, its share of the supply p.d. increases. The p.d. across the other resistor decreases. The two p.d.s always add up to the supply p.d. Extended
  • LDR in a potential divider: in brighter light, the LDR’s resistance falls. So the p.d. across the LDR falls, and the p.d. across the fixed resistor rises. Extended
  • Thermistor in a potential divider: when the temperature rises, the thermistor’s resistance falls. So the p.d. across the thermistor falls, and the p.d. across the fixed resistor rises. Extended
  • This output p.d. can switch something on. For example, corridor lights can come on when it gets dark. Extended

Model

No model for this topic yet.

Equations

  • Sources in series

    E = E1 + E2 + E3 + …

    E = combined e.m.f. (V); E1, E2, E3 = e.m.f. of each source (V); all sources connected the same way round

  • Resistors in series

    R = R1 + R2 + R3 + …

    R = combined resistance (Ω); R1, R2, R3 = resistance of each resistor (Ω)

  • Two resistors in parallelExtended

    R = R1R2 / (R1 + R2)

    R = combined resistance (Ω); R1, R2 = resistance of each resistor (Ω); the same as 1 / R = 1 / R1 + 1 / R2

  • Currents at a junctionExtended

    sum of currents into a junction = sum of currents out of the junction

    for a supply current I that splits into two branches: I = I1 + I2 (A)

  • P.d.s in seriesExtended

    V = V1 + V2 + …

    V = total p.d. across the series components (V); V1, V2 = p.d. across each component (V)

  • P.d. across parallel branchesExtended

    V = V1 = V2

    V = p.d. across the parallel arrangement (V); V1, V2 = p.d. across each branch (V)

  • Potential dividerExtended

    R1 / R2 = V1 / V2

    R1, R2 = the two resistors in series (Ω); V1, V2 = p.d. across each of them (V)

Sources in series

Three cells, each of e.m.f. 1.5 V, are connected in series, all the same way round. Find the combined e.m.f.

  • E = E1 + E2 + E3
  • E = 1.5 + 1.5 + 1.5
  • E = 4.5 V

One cell is now turned round. Find the new combined e.m.f.

  • The reversed cell’s e.m.f. is subtracted.
  • E = 1.5 + 1.5 − 1.5 = 1.5 V

Resistors in series

Resistors of 4.0 Ω, 6.0 Ω and 10 Ω are connected in series to a 12 V supply. Find the combined resistance and the current.

  • R = R1 + R2 + R3
  • R = 4.0 + 6.0 + 10
  • R = 20 Ω
  • I = V / R = 12 ÷ 20 = 0.60 A
  • This current is the same at every point in the circuit.

Two resistors in parallel Extended

Resistors of 12 Ω and 4.0 Ω are connected in parallel. Find the combined resistance.

  • Given: R1 = 12 Ω, R2 = 4.0 Ω
  • R = R1R2 / (R1 + R2)
  • R = (12 × 4.0) ÷ (12 + 4.0) = 48 ÷ 16
  • R = 3.0 Ω
  • Check: 3.0 Ω is less than 4.0 Ω, the smaller resistor. It must be.

Currents at a junction Extended

The current from a supply is 0.90 A. It splits into two branches. The current in one branch is 0.35 A. Find the current in the other branch.

  • Current in = current out: 0.90 = 0.35 + I2
  • I2 = 0.90 − 0.35
  • I2 = 0.55 A

P.d.s in series Extended

A 4.0 Ω resistor and a 2.0 Ω resistor are in series with a 9.0 V battery. Find the p.d. across each resistor.

  • Combined resistance: R = 4.0 + 2.0 = 6.0 Ω
  • Current: I = V / R = 9.0 ÷ 6.0 = 1.5 A (the same in both resistors)
  • V1 = IR1 = 1.5 × 4.0 = 6.0 V
  • V2 = IR2 = 1.5 × 2.0 = 3.0 V
  • Check: V1 + V2 = 6.0 + 3.0 = 9.0 V, the battery p.d.

P.d. across parallel branches Extended

A 10 Ω resistor and a 30 Ω resistor are connected in parallel to a 6.0 V supply. Find the current in each branch and the current from the supply.

  • Each branch has the full supply p.d.: V1 = V2 = 6.0 V
  • I1 = 6.0 ÷ 10 = 0.60 A
  • I2 = 6.0 ÷ 30 = 0.20 A
  • Supply current = 0.60 + 0.20 = 0.80 A (junction rule)
  • Check: combined R = 6.0 ÷ 0.80 = 7.5 Ω, and (10 × 30) ÷ (10 + 30) = 7.5 Ω.

Potential divider Extended

A 2.0 kΩ resistor (R1) and a 4.0 kΩ resistor (R2) are in series with a 12 V supply. Find the p.d. across each.

  • R1 / R2 = 2.0 ÷ 4.0 = 1/2, so V1 / V2 = 1/2
  • The p.d.s must add up to 12 V. Share 12 V in the ratio 1 : 2.
  • V1 = 12 × 1/3 = 4.0 V
  • V2 = 12 × 2/3 = 8.0 V

R2 is replaced by an LDR. In bright light its resistance is 1.0 kΩ. Find the p.d. across the LDR.

  • R1 / R2 = 2.0 ÷ 1.0 = 2, so V1 = 2 × V2
  • Share 12 V in the ratio 2 : 1.
  • V2 = 12 × 1/3 = 4.0 V across the LDR. Brighter light, lower resistance, smaller p.d.

Common mistakes

  • Students write that current is “used up” by each lamp in a series circuit. / The mark scheme wants: the current is the same at every point in a series circuit. It is energy that is transferred in the lamps.
  • Students write that two resistors in parallel have a combined resistance larger than either one. / The mark scheme wants: the combined resistance is less than either resistor alone.
  • Students give only one advantage of lamps in parallel when the question asks for two. / The mark scheme wants separate points: full p.d. for each lamp, each switched on its own, the others stay on if one breaks.
  • Students add the resistances of parallel resistors: 12 Ω and 4.0 Ω give 16 Ω. / The mark scheme wants R = R1R2 / (R1 + R2) = 3.0 Ω. Extended
  • Students say the p.d. across each parallel branch is shared between the branches. / The mark scheme wants: each parallel branch has the same p.d. as the whole arrangement. Extended
  • Students say the LDR’s p.d. rises in bright light. / The mark scheme wants: brighter light → LDR resistance falls → p.d. across the LDR falls. Extended

Exam tips

  • Draw circuit diagrams with a ruler and the correct symbols. Leave no gaps in the wires. Put the ammeter in series and the voltmeter in parallel.
  • State the rule asked for in one sentence. Example: “The current is the same at all points in a series circuit.”
  • Calculate: for series, add resistances and e.m.f.s. Then use I = V / R for the current. Show each step.
  • Explain the junction rule with charge: charge is not made or lost at a junction, so current in = current out. Extended
  • Calculate parallel resistance with R = R1R2 / (R1 + R2). Then check that your answer is smaller than the smaller resistor. Extended
  • Describe a potential divider sensor as a chain: change in light or temperature → change in resistance → change in share of the p.d. → change in output p.d. Extended
  • A typical 1-mark answer: “The resistance of an NTC thermistor decreases as its temperature increases.”
  • A typical 2-mark answer to “Give two advantages of connecting lamps in parallel”: “Each lamp can be switched on and off on its own (1). If one lamp breaks, the others still light (1).”
  • A typical 3-mark answer to “Explain why the p.d. across the LDR falls when the room gets brighter”: “In brighter light, the resistance of the LDR falls (1). The supply p.d. is shared in the ratio of the resistances (1). So the LDR gets a smaller share of the p.d. (1).” Extended