31-33 - Charge, Current, Potential Difference and Resistance

31-33 - Charge, Current, Potential Difference and Resistance

Electric circuits become much easier when the first three quantities do three separate jobs. Current tells you how quickly charge is passing a point. Potential difference tells you how much energy is transferred for each coulomb of charge. Resistance tells you how much potential difference is needed for each ampere of current. Keep those roles separate and the equations stop feeling like a pile of letters.

Current As A Rate

Imagine choosing one fixed point in a wire and counting the charge that crosses it. If 3 C crosses in one second, the current is 3 A. If the same 3 C takes ten seconds, the current is much smaller. The physics idea is a rate, not a store of charge.

Electric current is the rate of flow of charged particles. The symbol for current is I, and its unit is the ampere, A.

The equation is:

I=ΔQΔtI = \frac{\Delta Q}{\Delta t}
SymbolMeaningUnit
Icurrentampere, A
ΔQcharge that passes a point during the intervalcoulomb, C
Δttime intervalsecond, s

The Δ symbol matters: it means a change or an amount over an interval. So ΔQ/Δt is charge transferred per second over that interval. One ampere is one coulomb per second, so 1 A = 1 C s^-1.

A common wrong model is that current is a substance moving around the circuit and being used up. The correct model is sharper: charge moves, current is the rate of that movement, and energy transfer is described by potential difference, not by "using up current".

For Edexcel, a state what is meant by current answer needs the phrase rate of flow of charge or charged particles. Saying "the flow of electrons" is incomplete at A Level because it misses the rate idea, and not every conductor uses electrons as the charge carriers.

Using The Current Equation

Use I = ΔQ/Δt when the question links charge, current and time. The decision point is the word rate: current compares an amount of charge with the time taken for that charge to pass.

Worked example:

During a short pulse, 18 C of charge passes a point in a wire in 6.0 s. Calculate the average current.

First choose the equation. The question gives charge and time, and asks for a rate of flow, so:

I=ΔQΔtI = \frac{\Delta Q}{\Delta t}

Now substitute with units:

I=18 C6.0 s=3.0 C s−1I = \frac{18\ C}{6.0\ s} = 3.0\ C\,s^{-1}

Since C s^-1 is the same as A:

I=3.0 AI = 3.0\ A

Interpretation: on average, 3.0 C of charge crosses the chosen point each second.

Two checks keep this calculation honest:

  1. Time must be in seconds. If the question says 2.0 min, use 120 s.
  2. A larger time for the same charge gives a smaller current, because the charge is passing more slowly.

Guided check:

If 24 C passes in 2.0 min, first convert the time: 2.0 min = 120 s. Then:

I=24120=0.20 AI = \frac{24}{120} = 0.20\ A

A common error is to divide by 2.0 and write 12 A. That answer treats minutes as seconds and is too large by a factor of 60.

Potential Difference As Energy Per Charge

Now keep the charge moving, but ask a different question: how much energy is transferred as each coulomb passes through a component?

Potential difference, often called p.d. or voltage, is energy transferred per unit charge. The symbol is V, and the unit is the volt, also written V. The equation is:

V=WQV = \frac{W}{Q}
SymbolMeaningUnit
Vpotential difference across a componentvolt, V
Wwork done or energy transferredjoule, J
Qcharge passing through the componentcoulomb, C

One volt is one joule per coulomb: 1 V = 1 J C^-1.

Worked example:

48 J of energy is transferred to a component when 12 C of charge passes through it. Calculate the potential difference across the component.

Decision point: this is not asking how quickly charge flows, so do not use I = ΔQ/Δt. It is asking for energy per charge, so use:

V=WQV = \frac{W}{Q}

Substitute:

V=48 J12 C=4.0 J C−1=4.0 VV = \frac{48\ J}{12\ C} = 4.0\ J\,C^{-1} = 4.0\ V

Interpretation: each coulomb of charge transfers 4.0 J of energy to the component.

This is where "current is used up" breaks. The charge does not disappear inside the component. Instead, energy is transferred from the electrical supply to the component as charge moves through it. Potential difference measures that energy transfer per coulomb.

Guided check:

If a lamp has a p.d. of 6.0 V and 0.20 C passes through it, rearrange V = W/Q to W = VQ.

W=6.0×0.20=1.2 JW = 6.0 \times 0.20 = 1.2\ J

So 0.20 C transfers 1.2 J, and each 1 C would transfer 6.0 J.

Resistance As V Per Ampere

Resistance connects the p.d. across a component to the current through it. It is defined by:

R=VIR = \frac{V}{I}
SymbolMeaningUnit
Rresistanceohm, Ω
Vpotential difference across the componentvolt, V
Icurrent through the componentampere, A

One ohm is one volt per ampere: 1 Ω = 1 V A^-1.

The wording "across" and "through" is not decorative. Potential difference is measured across the component. Current is measured through the component. If the values do not refer to the same component at the same time, R = V/I will not describe that component.

Worked example:

A resistor has a p.d. of 9.0 V across it and a current of 0.30 A through it. Calculate its resistance.

Decision point: the question gives p.d. and current, so use the definition of resistance:

R=VIR = \frac{V}{I}

Substitute:

R=9.0 V0.30 A=30 V A−1=30 ΩR = \frac{9.0\ V}{0.30\ A} = 30\ V\,A^{-1} = 30\ \Omega

Interpretation: this component needs 30 V across it for each ampere of current through it, at this operating point.

Do not read resistance as "how much current is used up". A larger resistance means that, for the same p.d., the current would be smaller. Equally, for the same current, a larger resistance would require a larger p.d.

Guided check:

If V = 12 V and I = 2.0 A, then:

R=122.0=6.0 ΩR = \frac{12}{2.0} = 6.0\ \Omega

If the current were only 0.50 A for the same 12 V, the resistance would be:

R=120.50=24 ΩR = \frac{12}{0.50} = 24\ \Omega

Same p.d., smaller current, larger resistance.

Ohm's Law Is A Special Case

The equation R = V/I is a definition. You can calculate V/I for a component at a particular operating point. Ohm's law is a stronger statement: for an ohmic conductor at constant temperature, current is directly proportional to potential difference.

Directly proportional means:

  • if V doubles, I doubles
  • if V is zero, I is zero
  • the ratio V/I stays constant
  • an I against V graph would be a straight line through the origin

The constant-temperature condition matters because heating can change the resistance of a conductor. If the temperature changes while you take readings, a component may fail the proportionality test even if it would behave ohmically under controlled conditions.

Worked example:

A student records these readings for a metal resistor kept at constant temperature.

Potential difference V / VCurrent I / AV/I / Ω
2.00.1020
4.00.2020
6.00.3020

Determine whether the readings support Ohm's law.

Decision point 1: do not just say "current increases". Lots of components have current increase when p.d. increases. Ohm's law needs direct proportionality.

Decision point 2: use the data quantitatively. Here, doubling V from 2.0 V to 4.0 V doubles I from 0.10 A to 0.20 A, and V/I remains 20 Ω.

Decision point 3: include the condition. The conclusion is only an Ohm's-law conclusion because the resistor was kept at constant temperature.

A strong Edexcel determine response would say: the data support Ohm's law because I is directly proportional to V; for example, V/I is constant at 20 Ω for all three readings, and the resistor is at constant temperature.

Contrast this with:

V / VI / AV/I / Ω
2.00.1020
4.00.1625
6.00.2030

Current still increases, but not in direct proportion to p.d.; V/I is not constant. That is not enough evidence for Ohm's law.

Choosing Equations And Evidence

These three equations answer different questions:

If the question asks about...Use...Meaning check
charge flow rateI = ΔQ/Δtcoulombs per second
energy transferred per chargeV = W/Qjoules per coulomb
resistance of a componentR = V/Ivolts per ampere

Worked mixed example:

In 40 s, 16 C of charge passes through a component. The potential difference across the component is 6.0 V.

Calculate the current through the component and its resistance.

Step 1: use charge and time to find current.

I=ΔQΔt=1640=0.40 AI = \frac{\Delta Q}{\Delta t} = \frac{16}{40} = 0.40\ A

Step 2: use p.d. and current to find resistance.

R=VI=6.00.40=15 ΩR = \frac{V}{I} = \frac{6.0}{0.40} = 15\ \Omega

Decision point: do not put 16 C into R = V/I. Charge is not current. The charge first has to become a rate by dividing by time.

If the question also asked for energy transferred, then the p.d. equation would become useful:

W=VQ=6.0×16=96 JW = VQ = 6.0 \times 16 = 96\ J

That route has a different purpose: it finds total energy transferred, not current or resistance.

Practical/data transfer:

To measure resistance in a simple circuit, pair the readings correctly: measure the p.d. across the component and the current through it. A digital voltmeter and ammeter have finite resolution, so record readings to the displayed precision. If you are checking Ohm's law, keep the component at constant temperature by using small currents or switching off between readings, then look for a proportional pattern rather than trusting one reading.

Command-word transfer:

  • For calculate, show the equation, substitution, answer and unit.
  • For determine, use the numbers in the data to reach a quantitative conclusion.
  • For explain, connect the calculation or pattern to the physics reason.
  • For show that, carry extra precision in the working before rounding to the stated value.