Potential difference, energy and power

Section: Electricity | Syllabus: Cambridge AS Level Physics (9702)

Potential Difference

Potential difference (p.d.) between two points is defined as the energy transferred per unit charge passing between those points.

Potential Difference The energy transferred per unit charge: V = (W/Q)
where V = potential difference (V), W = energy transferred (J), Q = charge (C)

Potential difference is measured in volts (V). One volt is the p.d. between two points when one joule of energy is transferred per coulomb of charge passing between them.

Unit Relationship

1 V = 1 J C^-1 (one joule per coulomb)

Rearranging the Equation

The equation V = (W/Q) can be rearranged to find:

Energy transferred: W = VQ
Charge: Q = (W/V)

Since Q = It, we can also write:

W = VIt

Energy Transfer in Circuits

FIG 9.6: Simple Circuit Showing Current and Potential Difference

Show a cell connected in series with an ammeter (A) and a lamp. A voltmeter (V) is connected in parallel across the lamp to measure the potential difference.

As charge moves around a circuit, energy is transferred:

In the power supply: Work is done ON the charges, increasing their electrical potential energy. Chemical energy (battery) or other forms are converted to electrical energy.
In circuit components: Work is done BY the charges, transferring electrical energy to other forms (thermal, light, kinetic, etc.).

Energy Conservation

The total energy supplied by the source equals the total energy transferred in all components (conservation of energy).

Worked Example: Energy and Charge

Question: What is the potential difference across a motor when 30 C of charge passing through it transfers 90 J of energy?

Solution V = (W/Q) = (90/30) = 3 V Worked Example: Calculating Charge

Question: An electric motor runs from a 9 V battery and does 180 J of work. What charge passes through the motor?

Solution Q = (W/V) = (180/9) = 20 C

Electrical Power

Power is the rate of energy transfer (or rate of doing work).

Power P = (W/t)
where P = power (W), W = energy transferred (J), t = time (s)

Power is measured in watts (W). One watt equals one joule per second (1 W = 1 J s^-1).

Electrical Power Equations

For electrical circuits, power can be expressed in several equivalent forms:

Equation	Form	Useful When You Know
P = VI	Primary form	Voltage and current
P = I^2R	Derived using V = IR	Current and resistance
P = (V^2/R)	Derived using I = (V/R)	Voltage and resistance

Derivation of Power Equations

Deriving P = VI

Starting from the definitions of potential difference and power:

V = (W/Q) so W = VQ P = (W/t) = (VQ/t) Since I = (Q/t), we have (Q/t) = I Therefore: P = VI

Deriving P = I²R

Starting from P = VI and substituting V = IR:

P = VI = (IR) × I = I^2R

Deriving P = V²/R

Starting from P = VI and substituting I = (V/R):

P = VI = V × (V/R) = (V^2/R) Exam Tip

You must be able to derive these equations from P = VI and Ohm's law. Examiners often ask for derivations.

Interpreting Power Ratings

Electrical appliances have power ratings that indicate the rate at which they transfer energy under normal operating conditions.

For example, a "650 W" heater transfers 650 joules of electrical energy to thermal energy every second when operating at its rated voltage.

Worked Example: Power Calculation

Question: A lamp rated at 3.0 V passes a current of 500 mA. What is its power output?

Solution Convert current: I = 500 mA = 0.50 A P = VI = 3.0 × 0.50 = 1.5 W Worked Example: Finding Current from Power

Question: An electric shower has a power output of 7.7 kW and operates at 230 V. Calculate the current.

Solution P = VI so I = (P/V) I = (7700/230) = 33 A (to 2 s.f.)

Energy Transferred by Electrical Devices

The energy transferred by an electrical device can be calculated using:

Energy Equations W = Pt = VIt = I^2Rt = (V^2t/R) Worked Example: Energy Calculation

Question: A 2.0 kW kettle is used for 3.0 minutes. Calculate the energy transferred.

Solution Convert time: t = 3.0 × 60 = 180 s W = Pt = 2000 × 180 = 360\,000 J = 360 kJ Worked Example: Combined Calculation

Question: A 12 V power supply is connected to a lamp with resistance 6.0 Ω. Calculate (a) the current, (b) the power, and (c) the energy transferred in 5.0 minutes.

Solution

(a) Current:

I = (V/R) = (12/6.0) = 2.0 A

(b) Power:

P = VI = 12 × 2.0 = 24 W

Or: P = (V^2/R) = (144/6.0) = 24 W

(c) Energy transferred:

W = Pt = 24 × (5.0 × 60) = 24 × 300 = 7200 J = 7.2 kJ

The Kilowatt-Hour (kWh)

For practical purposes, electrical energy is often measured in kilowatt-hours (kWh) rather than joules.

Kilowatt-Hour 1 kWh = energy transferred by a 1 kW device operating for 1 hour
1 kWh = 1000 W × 3600 s = 3.6 × 10^6 J = 3.6 MJ Worked Example: Energy in kWh

Question: A 650 W microwave is used for 4 minutes. Calculate the energy transferred in (a) joules and (b) kWh.

Solution

(a) In joules:

W = Pt = 650 × (4 × 60) = 650 × 240 = 156\,000 J

(b) In kWh:

Power = 0.65 kW, Time = (4/60) hours = (1/15) hours Energy = 0.65 × (1/15) = 0.043 kWh

Common Exam Mistakes

Confusing V for volts and V for voltage: The italic V is the quantity (potential difference), while the upright V is the unit (volts).
Confusing W for watts and W for work: The italic W is work/energy in joules, while the upright W is the unit of power (watts).
Forgetting to convert units: Convert kW to W, minutes to seconds, mA to A before substituting.
Using wrong power formula: Choose the equation based on which quantities you know. Don't use P = I^2R if you don't know R.
Squaring incorrectly: In P = I^2R, only I is squared, not R. In P = (V^2/R), only V is squared.

Quick Concept Check

Define potential difference and state its SI unit.
A charge of 50 C passes through a heater, transferring 6000 J of energy. Calculate the p.d. across the heater.
Derive the equation P = I^2R starting from P = VI.
A 60 W lamp operates at 12 V. Calculate the current and the resistance of the lamp.
Calculate the energy transferred when a 2.5 kW heater runs for 20 minutes.
Convert 500 kJ to kWh.
A kettle is rated at 230 V, 2.0 kW. Calculate: (a) the current, (b) the resistance of the heating element, (c) the energy used to boil water for 2.5 minutes.

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Potential difference, energy and power