SPH3U Voltage, Resistance, Ohm's Law and Electrical Power: Notes and Practice

Key things to remember

1A volt is a joule per coulomb:

V

is the energy given to each unit of charge.

R = \frac{V}{I}

, rearranged as

V = I R

I = \frac{V}{R}

3An ohmic component has constant resistance and gives a straight line through the origin on a

V

–

I

graph.

4Resistance rises with length and with temperature, and falls with cross-sectional area.

P = V I

— volts times amperes gives watts, because

\frac{J}{C} \times \frac{C}{s} = \frac{J}{s}

P = I^{2} R

means power loss grows with the square of the current.

1 kWh = 3.6 \times 1 0^{6} J

, and it is a unit of energy, not power.

E (kWh) = \frac{P (W) \times t (h)}{1000}

, then multiply by the rate for the cost.

Three quantities describe any circuit. Current is how much charge flows. Potential difference is how much energy each coulomb carries. Resistance is how strongly the circuit opposes the flow. Every question in this unit is some combination of the three.

Quantity	Symbol	Unit	What it measures
Current	$I$	ampere (A)	Rate of charge flow — coulombs per second
Potential difference	$V$	volt (V)	Energy given to each coulomb — joules per coulomb
Resistance	$R$	ohm ( $Ω$ )	Opposition to current flow
Power	$P$	watt (W)	Rate of energy conversion — joules per second

A volt is a joule per coulomb

This is the definition worth holding on to. A

9 V

battery gives each coulomb of charge

9

joules of energy. Voltage is not 'how much electricity' — it is energy per unit charge, which is why a

9 V

battery and a

9 V

power supply differ in how much current they can deliver but not in the push each coulomb receives.

The water analogy helps as long as you do not lean on it too hard. Voltage is like the pressure difference driving water along a pipe, current is like the flow rate, and resistance is like the narrowness of the pipe. Narrow the pipe at fixed pressure and the flow drops — which is exactly Ohm's law.

Ohm's law

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

One relationship, three rearrangements. The formula sheet gives the first form.

Fig 1Cover the quantity you want and the triangle shows the arrangement.

The ohm is defined by this relationship: one ohm is the resistance that allows one ampere when one volt is applied. A component that obeys Ohm's law — one whose resistance stays constant as voltage changes — is called ohmic, and gives a straight line through the origin on a graph of $V$ against $I$ . Filament lamps are famously non-ohmic, because they heat up and their resistance rises with temperature.

Material — copper conducts far better than nichrome, which is why heating elements are nichrome and wires are copper.
Length — a longer conductor has more resistance, in direct proportion.
Cross-sectional area — a thicker conductor has less resistance, which is why high-current circuits need thick cable.
Temperature — in metals, resistance rises as temperature rises, because vibrating atoms obstruct the electrons more.

Electrical power

P = V I

Volts times amperes gives watts. Energy per coulomb times coulombs per second is energy per second.

The units make this one self-explaining: $V \times A = \frac{J}{C} \times \frac{C}{s} = \frac{J}{s} = W$ . Combining it with Ohm's law gives two more useful forms, $P = I^{2} R$ and $P = \frac{V ^{2}}{R}$ , though only $P = V I$ is on your formula sheet.

Why thin wires get hot

P = I^{2} R

says power loss grows with the square of the current. Double the current through a wire and it dissipates four times the heat. This is why an undersized extension cord warms up under load, why fuses work, and — on a much larger scale — why the grid transmits at very high voltage so that the current, and therefore the

I^{2} R

losses, stay small.

Energy and the hydro bill

Power is a rate. Energy is power multiplied by time, and for electricity the practical unit is the kilowatt-hour (kWh) — one kilowatt drawn for one hour. It is a unit of energy, not power, despite the name containing 'watt'.

E (kWh) = \frac{P (W) \times t (h)}{1000} cost = E \times rate

Divide by 1000 to convert watts to kilowatts. Ontario rates run roughly 7 to 18 ¢/kWh depending on time of use.

Appliance	Typical power	Energy in 1 hour
LED bulb	$10 W$	$0.010 kWh$
Laptop	$60 W$	$0.060 kWh$
Refrigerator (running)	$150 W$	$0.15 kWh$
Microwave	$1200 W$	$1.2 kWh$
Electric kettle	$1500 W$	$1.5 kWh$
Clothes dryer	$3000 W$	$3.0 kWh$

One kilowatt-hour in joules

1 kWh = 1000 W \times 3600 s = 3.6 \times 1 0^{6} J .

Worth memorizing, because questions switch between kilowatt-hours and joules without warning.

Time-of-use pricing

Ontario prices electricity by time of day, because demand peaks in the late afternoon and the most expensive, least clean generation runs to meet it. Shifting the dryer and dishwasher to overnight costs less and puts less strain on the grid — the physics and the economics point the same way here.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

12 V

battery drives

0.50 A

through a resistor. Find the resistance and the power dissipated.

Example 2

A hair dryer rated at

1500 W

runs on a

120 V

supply. Find the current it draws and its resistance.

Example 3

1200 W

microwave is used for

45 minutes

a day. At

12 ¢/kWh

, what does it cost to run for 30 days?

Example 4

How many joules does a

60 W

lamp use in

2.0 hours

? Express the answer in both joules and kilowatt-hours.

Where marks get lost

Treating the kilowatt-hour as a unit of power. It is energy — a kilowatt sustained for an hour.

Forgetting to divide watts by 1000 before multiplying by hours to get kilowatt-hours.

Using seconds with kilowatt-hours, or hours with joules. Match the units to the quantity you want.

Rearranging Ohm's law wrongly. Check with units: volts divided by amperes gives ohms.

Assuming every component is ohmic. A filament lamp's resistance rises sharply as it heats.

Thinking a higher-resistance appliance draws more power at fixed voltage. At constant

V

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

— higher resistance means less power.

Ignoring that

P = I^{2} R

is a square law, so doubling the current quadruples the heating.

Key things to remember

Worked examples

Where marks get lost

Flashcards 10

Practice