Potential difference and power

AS · 10 min

A current on its own does nothing useful: what matters is the energy the charge carries from the supply to the components. Potential difference measures how much energy each coulomb of charge transfers in a component, and power measures how fast that energy is transferred. This topic gives you $V = W/Q$ and the three power equations $P = VI$ , $P = I^2R$ and $P = V^2/R$ , which appear in almost every electricity question on Papers 1 and 2.

Energy and charge

Think of charge carriers as delivery vans. The supply loads each coulomb of charge with energy; as the charge passes through a lamp, a heater or a motor, it unloads energy there, and the component converts it into light, thermal energy or kinetic energy. The charge itself is not used up: the same coulombs go round and round, and the current is the same before and after the lamp.

So the useful question about a component is: how much energy does each coulomb transfer as it passes through? That is the potential difference.

Definition

The potential difference (p.d.) across a component is the energy transferred per unit charge (from electrical energy to other forms) as charge passes through the component:

V = \frac{W}{Q}

Definition

One volt is one joule per coulomb: $1\ \text{V} = 1\ \text{J C}^{-1}$ . A p.d. of $1\ \text{V}$ across a component means each coulomb passing through transfers $1\ \text{J}$ of energy to the component.

In SI base units, $1\ \text{V} = 1\ \text{J C}^{-1} = 1\ \dfrac{\text{kg m}^2\,\text{s}^{-2}}{\text{A s}} = 1\ \text{kg m}^2\,\text{s}^{-3}\,\text{A}^{-1}$ .

"Voltage" is a common everyday word for p.d. Use "potential difference" in written answers.

Key result

V = \frac{W}{Q} \qquad \text{so} \qquad W = VQ = VIt

$W$ : energy transferred (J); $Q$ : charge (C); $V$ : p.d. (V); $I$ : current (A); $t$ : time (s).

Measuring p.d.

A voltmeter compares the energy per unit charge at two points, so it is connected across a component (in parallel with it), with one lead on each side. An ideal voltmeter has infinite resistance so that no current is diverted through it. The p.d. across a length of connecting wire is taken as zero: ideal connecting wires transfer no energy.

Watch out

Potential difference is across a component; current is through a component. "The p.d. through the lamp" or "the current across the resistor" is wrong and loses credit in written answers. The charge flows through; the energy per unit charge is compared across.

Energy transferred by charge

A motor is connected to a $12\ \text{V}$ supply. (a) Calculate the energy transferred when $30\ \text{C}$ of charge passes through the motor. (b) The current in the motor is $2.5\ \text{A}$ . Calculate the energy transferred in $1.0$ minute.

Solution

(a)

W = VQ = 12 \times 30 = 360\ \text{J}

(b) First find the charge, $Q = It = 2.5 \times 60 = 150\ \text{C}$ , then

W = VQ = 12 \times 150 = 1800\ \text{J} = 1.8 \times 10^{3}\ \text{J}

or in one step, $W = VIt = 12 \times 2.5 \times 60 = 1.8 \times 10^{3}\ \text{J}$ .

Electrical power

Power is the rate of energy transfer, or the energy transferred per unit time:

P = \frac{W}{t}

Substitute $W = VQ$ and then $Q/t = I$ :

P = \frac{VQ}{t} = V \times \frac{Q}{t} = VI

The resistance of a component is $R = V/I$ (defined in Resistance and I–V characteristics). Substituting $V = IR$ or $I = V/R$ into $P = VI$ gives two more forms.

Key result

P = VI \qquad P = I^2R \qquad P = \frac{V^2}{R}

All three give the same answer for the same component; choose the one whose quantities you know. Power is measured in watts: $1\ \text{W} = 1\ \text{J s}^{-1} = 1\ \text{V A}$ .

Which form to use

$P = VI$ works for any component, including lamps, motors and diodes.
$P = I^2R$ is the natural choice when the current is fixed or known, for example in series components (same current) or power wasted in cables.
$P = V^2/R$ is the natural choice when the p.d. is fixed, for example for components in parallel (same p.d.) or appliances on the mains.

The forms $P = I^2R$ and $P = V^2/R$ appear to say opposite things: one says power increases with $R$ , the other that it decreases with $R$ . Both are right, under different conditions. For resistors in series (same $I$ ), the larger resistor dissipates more power. For resistors in parallel (same $V$ ), the smaller resistor dissipates more power.

Watch out

A power rating such as " $60\ \text{W},\ 230\ \text{V}$ " tells you the power when the stated p.d. is applied. On any other p.d. the power is different. If the resistance can be assumed constant, find $R = V^2/P$ from the rating, then use it with the new p.d. Do not assume the power stays at $60\ \text{W}$ .

A kettle

An electric kettle is rated at $2.8\ \text{kW}$ when connected to a $230\ \text{V}$ supply. Calculate (a) the current, (b) the resistance of the heating element, (c) the energy transferred in $3.0$ minutes.

Solution

(a)

I = \frac{P}{V} = \frac{2800}{230} = 12.2\ \text{A}

(b)

R = \frac{V^2}{P} = \frac{230^2}{2800} = 18.9\ \Omega

(c)

W = Pt = 2800 \times 180 = 5.04 \times 10^{5}\ \text{J}

A lamp on the wrong supply

A lamp is rated $6.0\ \text{V},\ 3.0\ \text{W}$ . (a) Calculate the current in the lamp and its resistance at normal brightness. (b) The lamp is connected to a $4.5\ \text{V}$ supply. Assuming its resistance does not change, calculate the power. (c) Explain whether the actual power will be more or less than your answer to (b).

Solution

(a)

I = \frac{P}{V} = \frac{3.0}{6.0} = 0.50\ \text{A}, \qquad R = \frac{V}{I} = \frac{6.0}{0.50} = 12\ \Omega

(b)

P = \frac{V^2}{R} = \frac{4.5^2}{12} = 1.7\ \text{W}

(c) At the lower p.d. the current is smaller, so the filament is cooler. A cooler metal filament has a lower resistance than $12\ \Omega$ . With a lower $R$ and the same $V$ , $P = V^2/R$ is larger, so the actual power is more than $1.7\ \text{W}$ (though still less than $3.0\ \text{W}$ ).

Why power is transmitted at high voltage

Power lost as thermal energy in a cable of resistance $R$ is $I^2R$ . For a fixed power $P = VI$ delivered, raising the transmission p.d. reduces the current in the same proportion, and the loss falls with the square of the current. This is why the National Grid and similar networks transmit at hundreds of kilovolts.

Cable losses

A workshop needs $20\ \text{kW}$ . The cable supplying it has a total resistance of $0.50\ \Omega$ . Calculate the power wasted in the cable if the power is delivered (a) at $230\ \text{V}$ , (b) at $2300\ \text{V}$ . Comment on the results.

Solution

(a) $I = P/V = 20\,000/230 = 87.0\ \text{A}$ .

P_{\text{loss}} = I^2R = 87.0^2 \times 0.50 = 3.8 \times 10^{3}\ \text{W}

(b) $I = 20\,000/2300 = 8.70\ \text{A}$ .

P_{\text{loss}} = 8.70^2 \times 0.50 = 38\ \text{W}

Increasing the p.d. by a factor of 10 reduces the current by a factor of 10 and the power loss by a factor of $10^2 = 100$ . At $230\ \text{V}$ almost a fifth of the power is wasted; at $2300\ \text{V}$ the loss is negligible. (Here "delivered at" is taken to mean the current needed for $20\ \text{kW}$ at that p.d.; the p.d. dropped across the cable is ignored.)

Two heaters in series

Two heaters are designed for a $230\ \text{V}$ supply: X is rated $1.0\ \text{kW}$ and Y is rated $2.0\ \text{kW}$ . They are connected in series across a $230\ \text{V}$ supply. Assuming their resistances are constant, calculate (a) the resistance of each heater, (b) the current, (c) the power of each heater, (d) the total power. Explain why the "more powerful" heater now gives out less power.

Solution

(a)

R_X = \frac{V^2}{P} = \frac{230^2}{1000} = 52.9\ \Omega, \qquad R_Y = \frac{230^2}{2000} = 26.5\ \Omega

(b) In series the resistances add (see Resistors in series and parallel):

I = \frac{230}{52.9 + 26.45} = \frac{230}{79.35} = 2.90\ \text{A}

(c)

P_X = I^2R_X = 2.90^2 \times 52.9 = 444\ \text{W}, \qquad P_Y = I^2R_Y = 2.90^2 \times 26.45 = 222\ \text{W}

(d) $P = 444 + 222 = 667\ \text{W}$ (check: $V^2/R_{\text{total}} = 230^2/79.35 = 667\ \text{W}$ ).

In series both heaters carry the same current, so $P = I^2R$ shows that the heater with the larger resistance (X, the $1.0\ \text{kW}$ one) dissipates more power. Y was more powerful only when it had the full $230\ \text{V}$ across it. The total power is less than either rating because the total resistance is larger than either resistance.

Tip

Electricity bills use the kilowatt-hour, the energy transferred by a $1\ \text{kW}$ appliance in $1$ hour: $1\ \text{kW h} = 1000 \times 3600 = 3.6 \times 10^{6}\ \text{J}$ . It is not part of the 9702 syllabus, but it can appear in context in a question, where it will be explained.

Exam tip

"Define potential difference" needs energy transferred per unit charge. "Energy per unit charge" alone may lose the mark if "transferred" is missing; "voltage per unit current" or "force per unit charge" score nothing.
When a question gives a power rating and a different supply p.d., examiners expect $R = V^2/P$ first. Say "assuming the resistance is constant".
Show the equation you use before substituting. In multi-step calculations keep at least three significant figures in intermediate values, then round the final answer to the precision of the data.
Units: kW to W, mA to A and minutes to seconds before substituting. A large share of lost marks in electricity questions come from unconverted prefixes.

Summary

The p.d. across a component is the energy transferred per unit charge: $V = W/Q$ . $1\ \text{V} = 1\ \text{J C}^{-1}$ .
Energy transferred: $W = VQ = VIt$ .
Power: $P = VI = I^2R = V^2/R$ , measured in watts.
Use $P = I^2R$ for series components (same current) and $P = V^2/R$ for parallel components (same p.d.).
A power rating applies only at the rated p.d.; find $R$ from it and recalculate on a different p.d.
Transmitting power at high p.d. means a small current and much smaller $I^2R$ losses.
P.d. is measured across a component with a voltmeter in parallel.

Practice

Question

A charge of $50\ \text{C}$ passes through a resistor and transfers $600\ \text{J}$ of energy. Calculate the p.d. across the resistor.
Show that the volt is equivalent to $\text{kg m}^2\,\text{s}^{-3}\,\text{A}^{-1}$ .
A $9.0\ \text{V}$ battery supplies a current of $0.15\ \text{A}$ for $2.0$ hours. Calculate the energy transferred.
A $230\ \text{V}$ hairdryer draws a current of $8.0\ \text{A}$ . Calculate its power and its resistance.
A $12\ \Omega$ resistor dissipates $3.0\ \text{W}$ . Calculate (a) the current, (b) the p.d. across it.
A lamp rated $230\ \text{V},\ 60\ \text{W}$ is used on a $110\ \text{V}$ supply. Assuming constant resistance, calculate the power.
A $4.0\ \Omega$ and a $6.0\ \Omega$ resistor are connected (a) in series to a $10\ \text{V}$ supply, (b) in parallel to a $10\ \text{V}$ supply. In each case state which resistor dissipates more power and calculate the power in each.
An electron is accelerated through a p.d. of $5.0\ \text{kV}$ . Calculate the energy it gains, in joules.
A battery-powered drill uses a $18\ \text{V}$ battery that can deliver a total charge of $7200\ \text{C}$ . The drill motor draws $6.0\ \text{A}$ and is $75\%$ efficient. Calculate (a) the total energy stored, (b) how long the battery lasts, (c) the useful mechanical power output.
A heating element is made of wire with resistance $R$ . It is connected to a supply of fixed p.d. $V$ and produces power $P$ . The wire is cut into two equal halves, which are connected in parallel across the same supply. Assuming resistances stay constant, find the new total power in terms of $P$ , and explain why this would be unsafe for a heater designed for power $P$ .

Answers

$V = W/Q = 600/50 = 12\ \text{V}$ .
$V = W/Q$ . Energy: $\text{J} = \text{N m} = \text{kg m s}^{-2} \times \text{m} = \text{kg m}^2\,\text{s}^{-2}$ . Charge: $\text{C} = \text{A s}$ . So $\text{V} = \dfrac{\text{kg m}^2\,\text{s}^{-2}}{\text{A s}} = \text{kg m}^2\,\text{s}^{-3}\,\text{A}^{-1}$ .
$W = VIt = 9.0 \times 0.15 \times 7200 = 9.7 \times 10^{3}\ \text{J}$ .
$P = VI = 230 \times 8.0 = 1840\ \text{W} \approx 1.8\ \text{kW}$ ; $R = V/I = 230/8.0 = 28.75 \approx 29\ \Omega$ .
(a) $I = \sqrt{P/R} = \sqrt{3.0/12} = 0.50\ \text{A}$ . (b) $V = \sqrt{PR} = \sqrt{36} = 6.0\ \text{V}$ (check: $VI = 3.0\ \text{W}$ ).
$R = 230^2/60 = 882\ \Omega$ ; $P = 110^2/882 = 13.7\ \text{W} \approx 14\ \text{W}$ . (Equivalently $P = 60 \times (110/230)^2$ .)
(a) Series: $I = 10/10 = 1.0\ \text{A}$ . $P_4 = 1.0^2 \times 4.0 = 4.0\ \text{W}$ , $P_6 = 6.0\ \text{W}$ : the $6.0\ \Omega$ resistor (larger $R$ , same $I$ ) dissipates more. (b) Parallel: each has $10\ \text{V}$ . $P_4 = 10^2/4.0 = 25\ \text{W}$ , $P_6 = 10^2/6.0 = 16.7\ \text{W}$ : the $4.0\ \Omega$ resistor (smaller $R$ , same $V$ ) dissipates more.
$W = VQ = 5.0 \times 10^{3} \times 1.60 \times 10^{-19} = 8.0 \times 10^{-16}\ \text{J}$ .
(a) $W = VQ = 18 \times 7200 = 1.3 \times 10^{5}\ \text{J}$ . (b) $t = Q/I = 7200/6.0 = 1200\ \text{s} = 20\ \text{min}$ . (c) Input power $= VI = 18 \times 6.0 = 108\ \text{W}$ ; useful output $= 0.75 \times 108 = 81\ \text{W}$ .
Each half has resistance $R/2$ . Two $R/2$ resistors in parallel have combined resistance $R/4$ . New power $= V^2/(R/4) = 4V^2/R = 4P$ . The current drawn from the supply would be four times larger, which could overheat the wiring and the element, melt insulation or blow a fuse; the heater is not designed to dissipate four times its rated power.

Energy and charge

Measuring p.d.

Electrical power

Which form to use

Why power is transmitted at high voltage

Summary

Practice

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