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Questions on circuit symbols and Kirchhoff's laws
Kirchhoff's first law is conservation of charge at a junction and his second law is conservation of energy around a closed loop. Before either come the standard circuit symbols, with the ammeter placed in series and the voltmeter across the component it measures. Current is not used up on the way round; energy is what a component takes.
25 original questions · 79 marks · the circuit symbols and Kirchhoff's laws notes · Electricity
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These questions cover both lessons on circuits and Kirchhoff's laws: Circuit symbols and Kirchhoff's laws and Series, parallel and power.
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening a mark scheme: the schemes award marks point by point, and the marks are easier to see when you have something of your own to compare against.
State Kirchhoff's first law and the conservation principle it expresses.
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The sum of the currents into a junction equals the sum of the currents out of it (1). This expresses conservation of charge (1).Two resistors of 20 Ω and 30 Ω are connected in series. Calculate the total resistance.
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In series resistances add: R = 20 + 30 (1)
R = 50 Ω (1)Two 20 Ω resistors are connected in parallel. Calculate the total resistance.
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1/R = 1/20 + 1/20 (1)
R = 10 Ω (1)Two 1.5 V cells are connected in series to form a battery. State the emf of the battery. State the emf if the two identical cells are instead connected in parallel, and give one advantage of the parallel arrangement.
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Series: emf = 1.5 + 1.5 = 3.0 V (1). Parallel: emf = 1.5 V, and the cells share the current between them, so each works less hard and the battery lasts longer (1).Explain why the combined resistance of two resistors in parallel is always less than the smaller of the two resistances.
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Adding a resistor in parallel adds an extra path for current without removing any (1). For the same pd the total current is therefore larger, so the combined resistance V/I is smaller than that of either branch alone (1).State which form of the power equation is most direct for comparing the powers dissipated in two resistors in series, and give the reason.
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P = I2R (1), because components in series share the same current (1).Draw a circuit diagram, using standard circuit symbols, in which a cell, a switch and a filament lamp are connected in series, an ammeter measures the current in the lamp, and a voltmeter measures the potential difference across the lamp.
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Standard symbols used, and recognisable, for the cell, the switch and the filament lamp (1). The ammeter is drawn in the series loop, so the same current passes through it and through the lamp (1). The voltmeter is drawn in parallel with the lamp, connected across it and not in the loop (1).A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel across a 12 V supply. Calculate (a) the total resistance and (b) the total current from the supply.
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(a) 1/R = 1/6.0 + 1/3.0 (1)
R = 2.0 Ω (1)
(b) I = V/R = 12/2.0 = 6.0 A (1)A 4.0 Ω and an 8.0 Ω resistor are connected in series across a 12 V supply. Calculate (a) the current in the circuit and (b) the potential difference across each resistor.
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(a) R = 4.0 + 8.0 = 12 Ω (1)
I = 12/12 = 1.0 A (1)
(b) V1 = IR1 = 4.0 V (1)
V2 = IR2 = 8.0 V (1)At a junction in a circuit, currents of 3.0 A and 2.0 A flow in, and a single wire carries the current out. Determine the current in the output wire, stating the law you use.
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By Kirchhoff's first law the current in equals the current out (1)
Iout = 3.0 + 2.0 = 5.0 A (1)A resistor of 5.0 Ω carries a current of 2.0 A. Calculate the power dissipated in it.
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P = I2R = 2.02 × 5.0 (1)
P = 20 W (1)An electric shower is rated 10.5 kW at 230 V. Calculate the current it draws and the resistance of its heating element.
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I = P/V = 10500/230 (1) = 46 A (1); R = V/I (or V2/P) = 5.0 Ω (1).A phone charger's packaging claims it delivers 38 kJ of energy in a 45 minute charge. In a test it supplies a steady 2.1 A at 5.0 V for the full 45 minutes. Deduce whether the claim is accurate.
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t = 45 × 60 = 2700 s (1); E = IVt = 2.1 × 5.0 × 2700 = 2.84 × 104 J (1); 28 kJ is well below the claimed 38 kJ, so the claim is not accurate (1).A battery of emf 24 V drives a current round a series loop containing three components. The pds across two of them are 9.0 V and 7.5 V. Determine the pd across the third component, stating the law you use.
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Kirchhoff's second law: round the loop the emf equals the sum of the pds (conservation of energy) (1); V = 24 − 9.0 − 7.5 = 7.5 V (1).A camping heater runs from a 12 V supply of negligible internal resistance and has two identical 4.6 Ω elements. Calculate the total power when the elements are connected (a) in series and (b) in parallel.
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(a) R = 4.6 + 4.6 = 9.2 Ω, and P = V2/R = 122/9.2 (1) = 16 W (1). (b) R = 4.6/2 = 2.3 Ω (1); P = 122/2.3 = 63 W (1).A 12 V charger is connected to a 9.0 V rechargeable battery so that the two emfs oppose each other, through a 3.0 Ω protective resistor. Both sources have negligible internal resistance. Calculate the current in the circuit, and state which way round the current passes through the battery.
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The two emfs oppose, so the loop is driven by their difference: 12 − 9.0 = 3.0 V (1). That is the only pd available to push a current through the resistor, so I = 3.0/3.0 = 1.0 A (1). The larger emf sets the direction, so the current enters the battery at its positive terminal, which is the direction that charges it (1).A 4.0 Ω resistor is connected in series with a parallel combination of a 6.0 Ω and a 3.0 Ω resistor, across a 12 V supply. Calculate (a) the total resistance, (b) the total current and (c) the current in each of the parallel resistors.
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Parallel part: 1/R = 1/6.0 + 1/3.0 = 0.50 Ω−1, so R = 2.0 Ω (1)
(a) Total = 4.0 + 2.0 = 6.0 Ω (1)
(b) I = 12/6.0 = 2.0 A (1)
(c) pd across the pair = I × 2.0 = 4.0 V, so I6 = 4.0/6.0 = 0.67 A and I3 = 1.33 A (1)A 60 W lamp is switched on for 5.0 minutes while connected to a 230 V supply. Calculate (a) the energy it transfers and (b) the charge that flows through it.
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(a) E = Pt = 60 × 300 (1)
E = 18000 J (1)
(b) E = QV, so Q = E/V = 18000/230 (1)
Q = 78.3 C (1)State Kirchhoff's second law and the conservation principle it expresses.
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Around any closed loop, the sum of the emfs equals the sum of the potential differences (1); equivalently, the sum of all pd changes around the loop is zero (1). This expresses conservation of energy (1).A car's two headlamps, each of resistance 6.0 Ω, are connected in parallel. The pair is in series with wiring of total resistance 0.40 Ω across the car's 12 V supply. The circuit is protected by a 5.0 A fuse. Deduce whether the fuse melts when both headlamps are on.
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Parallel pair: R = 6.0/2 = 3.0 Ω (1); total R = 3.0 + 0.40 = 3.4 Ω (1); I = 12/3.4 = 3.5 A (1); 3.5 A is below the 5.0 A rating, so the fuse does not melt (1).A 3.0 Ω resistor and a 9.0 Ω resistor are connected in series across a 6.0 V supply of negligible internal resistance. Calculate the power dissipated in each resistor, and state which runs hotter.
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I = 6.0/(3.0 + 9.0) = 0.50 A through both (1); P9 = I2R = 0.502 × 9.0 = 2.25 W (1); P3 = 0.502 × 3.0 = 0.75 W (1); the 9.0 Ω resistor runs hotter, since series components share the same current and P ∝ R (1).Two identical filament lamps are connected to a battery of negligible internal resistance, first in series and then in parallel. Explain fully why the lamps are brighter in parallel, and why the battery discharges sooner. A student claims that each lamp in series dissipates exactly one quarter of its parallel power. Comment on that claim.
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In series the two identical lamps share the supply pd, so each has half the supply pd across it (1). In parallel each lamp has the full supply pd across it, so each dissipates more power and glows brighter (1). The claim assumes a fixed resistance: with R constant, P = V2/R would give exactly a quarter (1). A filament lamp is not ohmic, since its resistance rises with temperature, and in series each filament runs cooler, so its resistance is lower than at full power (1). The lower resistance lets each series lamp draw more current than the constant-R model predicts, so the series power is more than a quarter of the parallel power and the total current in parallel is less than four times the series current; the claim is only a rough estimate (1). The parallel arrangement still draws much the larger total current, so charge leaves the battery at a higher rate and it transfers its stored energy sooner (1).A 9.0 V battery, two 1.5 V cells and a 12 Ω resistor are connected in a single loop. Both 1.5 V cells are connected so that they oppose the 9.0 V battery. Every source has negligible internal resistance. Calculate the net emf driving the loop and the current in the resistor, and state the potential difference across the resistor.
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Round the loop the emfs count with their signs: those met from the negative terminal to the positive one drive the current and add, and one met the other way round subtracts (1). Net emf = 9.0 − 1.5 − 1.5 = 6.0 V (1). I = ε/R = 6.0/12 = 0.50 A (1). The internal resistances are negligible, so the whole net emf appears across the resistor: 6.0 V (1).The circuit shown contains five ammeters, three of which are marked with their readings. Determine the readings of A3 and A5, naming the law you use and the conservation principle behind it.
A circuit containing five ammeters. Three readings are given; two are to be found. Mark scheme
Applying the junction rule where the lamp branch leaves the top wire: the current continuing along the top wire is 0.90 − 0.30 = 0.60 A, which then divides at the next junction between A3 and A4 (1)
A3 = 0.60 − 0.25 = 0.35 A (1)
A5 = 0.90 A, because the three branch currents recombine before A5, so it lies in the main line and reads the same as A1 (1)
This is Kirchhoff's first law, which follows from the conservation of charge: charge does not accumulate at a junction, so the current entering equals the current leaving (1)The circuit shown is driven by a supply of negligible internal resistance. Determine which of the three resistors dissipates the greatest power, and calculate that power.
Three resistors driven by a 12 V supply. Which one runs hottest is settled by where each sits. Mark scheme
The parallel pair combine as (12 × 6.0)/(12 + 6.0) (1)
That is 4.0 Ω for the pair, so the total resistance is 4.0 + 4.0 = 8.0 Ω and the supply current is I = 12/8.0 = 1.5 A (1)
The potential difference across the parallel pair is V = 1.5 × 4.0 = 6.0 V (1)
Choose the power form from the quantity each part shares: I2R for R1, which carries the whole current, and V2/R for the two branches, which share the 6.0 V (1)
R1 dissipates the greatest power, P = 1.52 × 4.0 = 9.0 W, against 3.0 W in R2 and 6.0 W in R3 (1)
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