9.2 - Potential difference and electrical power | Cambridge A-Level Physics

Interpret potential difference as energy transferred per unit charge, derive the electrical-power relations, and calculate energy transfers while keeping power, charge and time distinct.

Potential difference is energy per charge

A steady current passes through a lamp and transfers energy to it: does less charge leave the lamp than entered? No. Charge flow and energy transfer are different parts of the account. Potential difference tells you how much energy is transferred for each coulomb passing through a component.

A potential difference of $6.0\,\mathrm{V}$ across a component means an energy transfer of $6.0\,\mathrm{J}$ per coulomb. Transferring more charge through that same potential difference transfers more energy, rather than reducing the charge on each carrier.

Potential difference

The potential difference across a component is the energy transferred per unit charge passing through it.

The volt is a joule per coulomb. Use the potential difference across the component, not automatically the supply's labelled voltage: different components can share the total potential difference in a circuit.

Potential difference

V=\frac{W}{Q} \qquad W=VQ

$V$ is potential difference in volts (V), $W$ is energy transferred in joules (J), and $Q$ is charge transferred in coulombs (C). The product form applies when the p.d. is constant over that charge transfer, or when $V$ represents its appropriate average energy transfer per charge.

Not on the data sheet: learn it.

Worked example 1: energy transferred by a charge

Question. A charge of $18\,\mathrm{C}$ passes through a component with constant p.d. $4.5\,\mathrm{V}$ . Calculate the energy transferred.

Plan. Multiply joules per coulomb by the number of coulombs.

W=VQ=(4.5\,\mathrm{J\,C^{-1}})(18\,\mathrm{C})=81\,\mathrm{J}

Answer. $81\,\mathrm{J}$ .

Check. More than one coulomb passes, so the total energy is more than the energy per coulomb. The current cannot be determined unless the transfer time is known.

The transfer may increase internal energy, produce radiation, or do mechanical work, depending on the component. In a lamp, carriers transfer energy in the filament while charge continues through the circuit. A battery supplies energy to the circuit; the later e.m.f. lesson describes that transfer per unit charge around the complete circuit.

Watch out

Voltage is neither charge nor energy alone. A $12\,\mathrm{V}$ label does not mean “12 joules total” or “12 coulombs”: the amount of energy also depends on how much charge passes.

Separate the two quantities in a direct calculation.

Electrical power relations

To find how quickly a component transfers energy, combine the energy per coulomb with the coulombs passing per second. Their product is energy transferred per second: power.

Using $W=VQ$ and $I=Q/t$ for constant quantities,

P=\frac{W}{t}=V\frac{Q}{t}=VI

This explains the physical meaning of $VI$ : a high p.d. gives more energy per coulomb, while a high current carries more coulombs each second.

Resistance $R$ is the ratio of p.d. to current at an operating point, so $V=IR$ . The next lesson examines when this ratio stays constant. Substitution gives two alternative power expressions:

P=VI=I(IR)=I^{2}R \qquad P=VI=V\left(\frac{V}{R}\right)=\frac{V^{2}}{R}

Electrical power

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

$P$ is electrical power in W, $V$ is p.d. across the component in V, $I$ is current through it in A and $R=V/I$ is its resistance at that operating point in ohms ( $\Omega$ ). Use corresponding values for the same component and condition. If the resistance changes as a component warms, use its actual resistance for the state considered rather than assuming a fixed value.

Not on the data sheet: learn it.

Worked example 2: choosing the known pair

Question. A resistor has p.d. $12\,\mathrm{V}$ and current $0.30\,\mathrm{A}$ . Calculate its power.

Plan. Both $V$ and $I$ are known, so use $P=VI$ .

P=(12\,\mathrm{V})(0.30\,\mathrm{A})=3.6\,\mathrm{W}

Answer. $3.6\,\mathrm{W}$ .

Check. The resistor transfers $3.6\,\mathrm{J}$ each second, not $3.6\,\mathrm{C}$ each second.

For a fixed resistance, doubling current quadruples power, and doubling p.d. also quadruples power. But the comparisons must keep the named quantity fixed: at a fixed p.d., a larger resistance gives a smaller power because it reduces current. At a fixed current, a larger resistance requires more p.d. and transfers more power.

Exam-style example: the same resistor at a new voltage

Question. A resistor of constant resistance $48\,\Omega$ is connected first across $12\,\mathrm{V}$ , then across $18\,\mathrm{V}$ . Calculate its power at each voltage.

Plan. Use $P=V^{2}/R$ because voltage and resistance are known.

P_1=\frac{(12\,\mathrm{V})^{2}}{48\,\Omega}=3.0\,\mathrm{W} \qquad P_2=\frac{(18\,\mathrm{V})^{2}}{48\,\Omega}=6.75\,\mathrm{W}

Answer. $3.0\,\mathrm{W}$ and $6.8\,\mathrm{W}$ (2 s.f.).

Check. The voltage increased by a factor of 1.5, so power increased by $1.5^{2}=2.25$ , rather than by 1.5. The constant-resistance condition is essential.

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9.2 - Potential difference and electrical power

Potential difference is energy per charge

Electrical power relations

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