Current, Voltage & Resistance — Ohm's Law
Thomas Edison's Pearl Street power station (New York City, 4 September 1882) — the world's first commercial electricity utility: 110 V DC, copper conductors with resistance 0.5 Ω/km. At 100 A current: voltage drop = IR = 100 × 0.5 = 50 V/km. Edison designed his 110 V system so a 50 V drop over 1 km still left 60 V at the customer — a direct application of Ohm's law in system design.
A standard electric kettle is rated at 2400 W and is connected to the 240 V mains supply. A 100 W light globe is connected to the same supply.
Predict 1: Which draws more current from the supply — the kettle or the light globe? By approximately what factor?
Predict 2: If a wire has resistance, some energy is wasted as heat as current flows through it. What factors do you think would increase the resistance of a wire?
Warm-up — electric current is defined as:
Know
- Current: $I = Q/t$ (ampere = coulomb per second)
- Ohm's Law: $V = IR$ for ohmic resistors
- Factors affecting resistance of a conductor
- Difference between ohmic and non-ohmic devices
Understand
- Why resistance increases with temperature in metals
- Why $V$–$I$ graphs distinguish ohmic from non-ohmic devices
Can Do
- Calculate current, voltage, or resistance using Ohm's Law
- Interpret $V$–$I$ graphs to identify ohmic/non-ohmic behaviour
- Calculate charge transferred: $Q = It$
Core Content
Touch one terminal of a 9 V battery to a wire and nothing moves. Connect the other terminal too — and within microseconds, electrons throughout the entire wire begin drifting toward the positive terminal. You cannot see them move, but clip a small LED into the circuit and it glows instantly, because charge is flowing everywhere in the loop simultaneously, not just near the battery. This coordinated flow of charge is electric current.
$I = \dfrac{Q}{t}$ unit: ampere (A = C s⁻¹)
A current of 1 A means 1 coulomb of charge passes any cross-section of the conductor every second. The kettle in your kitchen draws about 10 A from the 240 V mains.
Electric current $I = Q/t$ (A = C s⁻¹) is the rate of charge flow. Conventional current flows from + to −; electron flow is − to + (opposite). Charge transferred: $Q = It$.
Pause — copy the highlighted definition and formula into your book before moving on.
A current of 3.0 A flows in a wire for 2.0 minutes. The total charge that passes any cross-section is:
We just saw that current $I = Q/t$ is the flow of charge. That raises a question: what opposes that flow, and how does voltage drive current through it? This card answers it → resistance and Ohm's Law $V = IR$.
Resistance is the property of a conductor that opposes the flow of current. It arises from collisions between conduction electrons and the lattice ions in the metal — each collision converts some electrical energy to heat (thermal energy).
$V = IR$ where $V$ = voltage (V), $I$ = current (A), $R$ = resistance (Ω)
Rearranged: $I = V/R$ and $R = V/I$
Factors Affecting Resistance
- Length: $R \propto L$ — longer wire → more collisions → higher resistance.
- Cross-sectional area: $R \propto 1/A$ — thicker wire → more pathways → lower resistance.
- Material: resistivity $\rho$ — $R = \rho L/A$.
- Temperature: in metals, higher temperature → more lattice vibration → higher resistance.
Ohmic vs Non-Ohmic Devices
- Ohmic: $R$ is constant (independent of $V$ or $I$). $V$–$I$ graph is a straight line through origin. Example: resistor at constant temperature.
- Non-ohmic: $R$ changes. $V$–$I$ graph is curved. Examples: filament light globe (resistance increases with temperature), diode (current only in one direction).
Ohm's Law: $V = IR$ (Ω = V A⁻¹). Resistance increases with length, decreases with cross-sectional area, increases with temperature (metals): $R = \rho L/A$. Ohmic devices show a straight $V$–$I$ line through the origin; non-ohmic devices show a curved graph.
Add the highlighted law and factors to your notes before the check below.
A filament light globe is an example of an ohmic device.
Doubling the length of a wire halves its resistance.
We just saw that Ohm's Law $V = IR$ applies to ohmic conductors with constant $R$. That raises a question: how do we identify from data whether a device is ohmic or non-ohmic? This card answers it → the shape of the $V$–$I$ graph reveals it.
A $V$–$I$ graph (voltage on $y$-axis, current on $x$-axis) reveals whether a device is ohmic:
- The gradient of the $V$–$I$ graph = resistance $R$ (in Ω)
- If the gradient is constant (straight line through origin), the device is ohmic
- If the gradient increases, resistance is increasing with current (e.g. filament globe heating up)
- If the gradient decreases, resistance is decreasing with current (e.g. thermistor cooling down... or for an LED/diode)
A toaster connected to a 240 V supply draws a current of 4.0 A. Find its resistance.
- $R = V/I = 240/4.0 = 60\ \Omega$
Answer: $R = 60\ \Omega$
On a $V$–$I$ graph (voltage on $y$, current on $x$), the gradient equals resistance $R$. A straight line through the origin indicates an ohmic device (constant $R$); a curved line indicates a non-ohmic device (varying $R$). On an $I$–$V$ graph the gradient equals conductance $1/R$.
Pause — write the highlighted graph rules into your book.
On a $V$–$I$ graph for a filament globe, the slope of the curve as current increases would:
Use $V = IR$ and $I = Q/t$ to solve:
- A 240 V electric kettle draws 8.0 A. Calculate its resistance.
- A resistor has resistance 120 Ω and is connected to a 12 V battery. Calculate the current through it.
- If the kettle in (1) runs for 90 seconds, calculate the total charge that flows through it.
Three statements correctly describe ohmic resistors. Pick the odd one out.
A student measured the current through a component at different voltages:
| V (V) | I (A) | R = V/I (Ω) |
|---|---|---|
| 2.0 | 0.20 | |
| 4.0 | 0.40 | |
| 6.0 | 0.60 | |
| 8.0 | 0.60 |
- Calculate the resistance at each voltage and fill in the third column in your response.
- Is this device ohmic or non-ohmic? Justify with reference to your calculations.
- What could explain the change in resistance at high voltage?
A 60 Ω resistor is connected to a 12 V battery. The current flowing through it is _____ A.
Current
$I = Q/t$ (A = C s⁻¹). Conventional current: + to −. Electrons: − to +.
Ohm's Law
$V = IR$. Ohmic: constant $R$, straight $V$–$I$ graph. Non-ohmic: varying $R$, curved graph.
Resistance factors
$R \propto L$; $R \propto 1/A$; $R = \rho L/A$. In metals: higher temperature → higher resistance.
V–I graph
Gradient of $V$–$I$ graph = $R$. Straight line through origin = ohmic. Curved = non-ohmic.
UnderstandBand 3(3 marks) 3. Explain, in terms of electron–lattice collisions, why the resistance of a metal wire increases as the temperature increases.
ApplyBand 4(3 marks) 4. A student measures the following V–I data for a resistor: V = 2 V → I = 0.5 A; V = 4 V → I = 1.0 A; V = 6 V → I = 1.5 A. (a) Calculate $V/I$ at each voltage and determine if the device is ohmic. (b) Find the resistance of the device.
EvaluateBand 6(4 marks) 5. A household circuit is protected by a 10 A fuse. A student replaces it with a 20 A fuse because the 10 A fuse kept blowing. Evaluate why the original fuse was blowing, why replacing it with a higher-rated fuse is dangerous, and what the correct solution would be.
Show all answers
Activity 1 — Model Answers
- $R = V/I = 240/8.0 = 30\ \Omega$
- $I = V/R = 12/120 = 0.10\ \text{A}$
- $Q = It = 8.0 \times 90 = 720\ \text{C}$
Activity 2 — Model Answers
- 2.0 V: $R = 10\ \Omega$; 4.0 V: $R = 10\ \Omega$; 6.0 V: $R = 10\ \Omega$; 8.0 V: $R = 13.3\ \Omega$
- Non-ohmic — resistance is constant at low voltages (10 Ω) but increases to 13.3 Ω at 8 V. The $V$–$I$ graph would show a straight line then a curve upward.
- The increased current at high voltage heats the filament. Higher temperature → more lattice vibrations → higher resistance. This is characteristic of a filament globe (non-ohmic behaviour).
Short Answer — Model Answers
Q3 (3 marks): At higher temperatures, the lattice ions in the metal vibrate more vigorously. Conduction electrons undergo more frequent and more energetic collisions with these vibrating lattice ions. Each collision transfers energy from the electron to the lattice (heat) and impedes the electron's progress. This increased collision rate means more opposition to charge flow — higher resistance.
Q4 (3 marks): (a) $V/I = 2/0.5 = 4\ \Omega$; $4/1.0 = 4\ \Omega$; $6/1.5 = 4\ \Omega$. The ratio is constant, so the device is ohmic. (b) $R = 4.0\ \Omega$.
Q5 (4 marks): The 10 A fuse was blowing because the circuit current exceeded 10 A — indicating the circuit was overloaded (too many appliances). A fuse melts to break the circuit before wires overheat and start a fire. Replacing it with a 20 A fuse allows current up to 20 A through wires rated only for 10 A — the wires can overheat, melt insulation, and cause a fire without the protection breaking the circuit. The correct solution is to reduce the load on the circuit by plugging fewer appliances in, or have an electrician install an additional circuit.
Five timed questions on current, voltage, and resistance.
⚔ Enter the arenaThomas Edison's Pearl Street power station (New York City, 4 September 1882) ran 110 V DC through copper conductors with resistance 0.5 Ω/km. At 100 A of load current the voltage drop was V = IR = 100 × 0.5 = 50 V/km — leaving customers 1 km away with only 60 V. Edison's engineers had to keep cable runs short precisely because of Ohm's Law. Today we use transformers to send power at high voltage (low current) so the IR drop is negligible — the very problem Pearl Street forced Edison to confront in 1882.
Now look back at your Think First answers: the kettle draws I = P/V = 2400/240 = 10 A; the globe draws 100/240 ≈ 0.42 A — the kettle draws about 24× more current. Resistance of a wire increases with greater length, smaller cross-section, and higher temperature — exactly the factors Pearl Street's engineers had to manage.