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Questions on current, charge and potential difference
Electric current is the rate of flow of charge, I = ΔQ/Δt, and potential difference is the energy transferred per unit charge, V = W/Q. One volt is one joule per coulomb. A charge accelerated from rest through a pd in a vacuum gains kinetic energy qV, which is what an electron gun exploits.
25 original questions · 74 marks · the current, charge and potential difference notes · Electricity
To print: the questions (PDF, 89 KB) · the worked answers (PDF, 85 KB)
These questions cover both lessons on current, charge and the direction problem: Current, charge and potential difference and Charge carriers, drift speed and resistance.
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.
A charge of 30 C flows past a point in 10 s. Calculate the current.
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I = ΔQ/Δt = 30/10 (1)
I = 3.0 A (1)Define the coulomb.
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One coulomb is the charge that passes a point when a current of one ampere flows for one second (1); that is, 1 C = 1 A s (1).12 J of work is done moving a charge of 4.0 C between two points. Calculate the potential difference.
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V = W/Q = 12/4.0 (1)
V = 3.0 V (1)A camera flash produces a current of 0.75 A that lasts for 2.4 ms. Calculate the charge that flows through the flash lamp.
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Use of Q = IΔt with t = 2.4 × 10−3 s (1); Q = 0.75 × 2.4 × 10−3 = 1.8 × 10−3 C (1).A phone battery is rated at 3200 mA h, meaning it can supply a current of 3200 mA for one hour. Calculate the charge the battery stores.
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I = 3.2 A and t = 3600 s (1); Q = It = 3.2 × 3600 = 1.15 × 104 C (1).An electron moving between two electrodes in a vacuum tube gains 8.0 × 10−19 J of energy. Calculate the potential difference between the electrodes (e = 1.60 × 10−19 C).
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V = W/Q (1); V = (8.0 × 10−19)/(1.60 × 10−19) = 5.0 V (1).Define the resistance of a component, and give the unit of resistance in terms of the volt and the ampere.
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Resistance is the potential difference across the component divided by the current through it, R = V/I (1). Its unit is the ohm, which is one volt per ampere (1).A current of 0.50 A flows for 2.0 minutes. Calculate (a) the charge that passes and (b) the number of electrons this represents (e = 1.60 × 10−19 C).
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(a) Q = It = 0.50 × 120 (1)
Q = 60 C (1)
(b) N = Q/e = 60/(1.60 × 10−19) = 3.75 × 1020 (1)A component has a potential difference of 6.0 V across it and carries a current of 0.25 A. Calculate its resistance.
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R = V/I = 6.0/0.25 (1)
R = 24 Ω (1)A charge of 8.0 C passes through a resistor with a potential difference of 12 V across it. Calculate the energy transferred in the resistor.
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W = QV = 8.0 × 12 (1)
W = 96 J (1)A steady current of 2.0 A flows in a lamp filament for 5.0 minutes. Calculate the charge that passes through the filament.
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Q = It = 2.0 × 300 (1)
Q = 600 C (1)The flexible lead to a kettle contains copper wire of diameter 1.2 mm carrying a current of 13 A. Copper has 8.5 × 1028 free electrons per cubic metre (e = 1.60 × 10−19 C). Calculate (a) the cross-sectional area of the wire and (b) the drift velocity of the electrons.
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(a) A = πd2/4 = π × (1.2 × 10−3)2/4 = 1.1 × 10−6 m2 (1). (b) Use of I = nAvq rearranged to v = I/nAq (1); v = 13/(8.5 × 1028 × 1.13 × 10−6 × 1.60 × 10−19) (1) = 8.5 × 10−4 m s−1 (1).A wire in a circuit tapers so that its cross-sectional area halves. State what happens to the drift velocity of the electrons where the wire narrows, and explain your answer using I = nAvq.
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The drift velocity doubles (1). The current is the same at every cross-section, because charge cannot accumulate at any point (conservation of charge) (1). In I = nAvq, n and q are unchanged, so halving A means v must double to carry the same current (1).A cyclist's rear lamp is powered by a 3.0 V battery and draws a current of 0.20 A. Calculate the energy transferred by the lamp during a 10 minute ride.
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t = 600 s (1); W = VIt = 3.0 × 0.20 × 600 (1) = 360 J (1).An electron starts from rest and is accelerated through a potential difference of 1.5 kV in a vacuum. Calculate its final speed. (e = 1.60 × 10−19 C; me = 9.11 × 10−31 kg.)
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There is nothing in a vacuum for the electron to hand its energy to, so all the work done on it becomes kinetic energy: eV = ½mv2 (1). v = √(2eV/m) = √(2 × 1.60 × 10−19 × 1500/(9.11 × 10−31)) (1) = 2.3 × 107 m s−1 (1).A current passes through copper sulfate solution between two electrodes. Name the charge carriers in the solution, state the direction in which each kind of carrier moves, and explain why the two flows add rather than cancel.
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The carriers are ions of both signs, positive and negative, which are free to move because the ionic compound has been dissolved (1). The positive ions drift towards the negative electrode and the negative ions towards the positive electrode (1). Negative charge moving one way is equivalent to positive charge moving the other way, so both flows count as conventional current in the same direction and their contributions add (1).A device draws a current of 3.0 A from a 230 V supply for 1.0 hour. Calculate (a) the charge that flows, (b) the energy transferred and (c) the number of electrons that pass (e = 1.60 × 10−19 C).
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(a) Q = It = 3.0 × 3600 (1)
Q = 10800 C (1)
(b) W = QV = 10800 × 230 = 2.48 × 106 J (1)
(c) N = Q/e = 6.75 × 1022 (1)Explain what is meant by conventional current and how it relates to the actual direction electrons move in a metal wire.
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Conventional current is defined as the direction in which positive charge would flow (1), from the positive terminal to the negative terminal in the external circuit (1). In a metal the charge carriers are electrons, which are negative (1) and therefore drift in the opposite direction to the conventional current (1).Explain why the convention for current direction was not changed once electrons were discovered to be the charge carriers in metals.
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The convention was fixed long before the electron was discovered (1), and all the established laws and circuit rules were written in terms of it (1). They work correctly as they are, so changing the convention would cause confusion for no benefit; the direction is simply treated as that of positive charge flow (1).An electric bicycle battery needs at least 2.0 × 105 C of charge to charge fully. It is connected for 8.0 hours overnight to a charger that supplies a steady current of 6.5 A. Deduce whether the battery charges fully.
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t = 8.0 × 3600 = 28800 s (1); Q = It = 6.5 × 28800 = 1.87 × 105 C (1); this is less than 2.0 × 105 C, so the battery does not charge fully (1).A copper track (8.5 × 1028 free electrons per m3) and a semiconductor channel (5.0 × 1019 charge carriers per m3) in a sensor have the same cross-sectional area and carry the same current. Determine the ratio of the drift velocity in the semiconductor to that in the copper.
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From I = nAvq, with I, A and q the same for both, v ∝ 1/n (1); ratio = nCu/nsemi = (8.5 × 1028)/(5.0 × 1019) (1) = 1.7 × 109 (1).In a metal wire the electrons drift at less than a millimetre per second, yet a lamp lights the instant the switch is closed and the current can be several amperes. Explain fully how both observations are consistent with the drift of electrons.
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In a metal the charge carriers are free electrons, drifting slowly along the wire (1). The current is I = nAvq, so its size depends on the carrier density n as well as the drift velocity v (1). n is enormous, around 1028 to 1029 per m3, so even a tiny v gives a current of several amperes (1). When the switch closes, an electric field is established around the whole circuit at close to the speed of light (1). Electrons everywhere in the circuit, including inside the lamp, begin to drift at almost the same instant (1). The lamp therefore lights immediately; no individual electron needs to travel from the switch to the lamp (1).A 9.0 V battery for a smoke alarm stores 1.9 × 104 C of charge. (a) Show that the energy stored is about 1.7 × 105 J. (b) In standby the alarm draws a steady current of 60 μA. The manufacturer claims a 10 year standby life. Determine whether the claim is justified (1 year = 3.15 × 107 s).
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(a) W = QV = 1.9 × 104 × 9.0 (1) = 1.71 × 105 J ≈ 1.7 × 105 J (1). (b) t = Q/I = (1.9 × 104)/(6.0 × 10−5) = 3.2 × 108 s (1); t = 10.1 years, so the claim is (just) justified (1).In the first stage of an accelerator a proton, of charge 1.60 × 10−19 C and mass 1.67 × 10−27 kg, starts from rest and crosses a potential difference of 25 kV in a vacuum. Calculate the kinetic energy the proton gains and its final speed, and state one assumption made in using ½mv2 for its kinetic energy.
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Work done on the charge W = QV = 1.60 × 10−19 × 25 000 = 4.0 × 10−15 J, and in a vacuum all of it becomes kinetic energy (1). ½mv2 = 4.0 × 10−15 J (1), so v = √(2 × 4.0 × 10−15/(1.67 × 10−27)) = 2.2 × 106 m s−1 (1). ½mv2 is a non-relativistic expression, so it assumes the final speed is far below the speed of light; here it is about 0.7% of c, so the assumption holds (1).A strip of doped semiconductor is 2.0 mm wide and 0.30 mm thick. It carries a current of 1.2 mA, and each charge carrier in it has a charge of 1.60 × 10−19 C and drifts at 3.5 m s−1. Calculate the number density of the charge carriers in the strip, and deduce whether the strip could be made of metal (copper has 8.5 × 1028 free electrons per m3).
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A = 2.0 × 10−3 × 0.30 × 10−3 = 6.0 × 10−7 m2 (1)
I = nAvq rearranged to n = I/(Avq) (1)
n = (1.2 × 10−3)/(6.0 × 10−7 × 3.5 × 1.60 × 10−19) = 3.6 × 1021 m−3 (1)
That is smaller than copper's free-electron density by a factor of about 2.4 × 107, so the strip cannot be a metal; a carrier density this low belongs to a semiconductor (1)
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