Resistance and I–V characteristics

AS · 13 min

Some components let current through easily; others resist it. Resistance measures how much p.d. is needed to drive each ampere through a component. This topic defines resistance, states Ohm's law, and shows the current–voltage ( $I$ – $V$ ) graphs of a metal at constant temperature, a filament lamp and a semiconductor diode: sketching these and explaining their shapes are some of the most reliable marks in Paper 2, and reading resistance from a curve is a favourite Paper 1 trap.

Resistance

If you apply a p.d. across a component, a current flows. The bigger the resistance, the smaller the current for a given p.d.

Definition

The resistance of a component is the ratio of the potential difference across it to the current in it:

R = \frac{V}{I}

Definition

One ohm ( $\Omega$ ) is the resistance of a component in which a p.d. of one volt drives a current of one ampere: $1\ \Omega = 1\ \text{V A}^{-1}$ .

Key result

V = IR

$V$ in volts, $I$ in amperes, $R$ in ohms. This is the definition of resistance rearranged: it is true for every component at every instant, whether or not the component obeys Ohm's law.

What causes resistance

In a metal, the free electrons drift through a lattice of positive ions. They keep colliding with the ions (which vibrate about fixed positions), transferring some of their energy to the lattice. That is why a current heats a wire. Anything that makes collisions more frequent, such as hotter, more strongly vibrating ions, increases the resistance. Anything that increases the number of charge carriers available, as in a semiconductor that is warmed or illuminated, reduces it.

Ohm's law

For many conductors the current is proportional to the p.d., so $R$ is constant. This is a special property, not a general law of electricity.

Definition

Ohm's law: the current in a conductor is directly proportional to the potential difference across it, provided that physical conditions (such as temperature) remain constant.

A component that obeys Ohm's law is called ohmic. A metal wire or a fixed resistor at constant temperature is ohmic. A filament lamp, a diode and a thermistor are non-ohmic.

Watch out

$V = IR$ is not Ohm's law. $V = IR$ defines resistance and applies to every component. Ohm's law is the extra statement that $I \propto V$ (that is, $R$ is constant) for certain conductors at constant temperature. Examiners regularly ask for a statement of Ohm's law and do not accept " $V = IR$ ".

Measuring an I–V characteristic

To plot the $I$ – $V$ characteristic of a component, you vary the p.d. across it and measure the current for each value.

Circuit for an I–V characteristic. The variable resistor changes the current; the ammeter is in series with the component and the voltmeter is connected across the component only.

Connect the component in series with an ammeter and a variable resistor (rheostat) to a d.c. supply. Connect a voltmeter across the component only.
Adjust the variable resistor to change the p.d. across the component. Record $V$ and $I$ for at least six values spread over the range.
Reverse the connections to the supply (or to the component) and repeat to obtain negative values of $V$ and $I$ .
Plot $I$ on the $y$ -axis against $V$ on the $x$ -axis.

A variable resistor in series cannot reduce the p.d. to zero. To cover the whole range from $0\ \text{V}$ , use the variable resistor as a potential divider instead (see Potential dividers).

Three characteristics to know

Metallic conductor at constant temperature

y = 0.4 x

The graph of $I$ against $V$ is a straight line through the origin: $I \propto V$ , so the resistance is constant. The line has the same gradient for negative p.d.s, because reversing the p.d. simply reverses the current. The gradient of this line is $I/V = 1/R$ , so a steeper line means a smaller resistance.

Filament lamp

y = x / (10 + 2 abs(x))

The graph is an S-shaped curve through the origin, symmetric for negative values. Near the origin it is almost straight. As the p.d. increases, the curve bends towards the $V$ -axis: each extra volt produces a smaller increase in current.

Why the lamp's resistance increases

As the current increases, the filament's temperature increases. The metal ions vibrate with greater amplitude, so the free electrons collide with them more often. This makes it harder for charge to flow, so the resistance increases. A larger $V/I$ at higher current shows on the graph as the curve bending towards the $V$ -axis.

Semiconductor diode

y = 0.00001 (e^(25 x) - 1)

Current ( $I$ , in mA) against p.d. ( $V$ , in V) for a silicon diode.

Reverse bias (negative $V$ ): the current is effectively zero. The diode has a very high resistance and does not conduct.
Forward bias, below about $0.6\ \text{V}$ for a silicon diode: very little current.
Forward bias, above about $0.6$ to $0.7\ \text{V}$ : the current rises very steeply. The diode now has a low resistance and the p.d. across it stays close to $0.7\ \text{V}$ over a wide range of currents.

So a diode lets current through in one direction only, the direction the triangle in its symbol points. This makes it useful to protect circuits against a supply connected the wrong way round, and to rectify alternating current (an A Level topic). A light-emitting diode (LED) has the same shape of characteristic with a larger switch-on p.d. (roughly $1.5$ to $3\ \text{V}$ depending on its colour).

Tip

The thermistor's $I$ – $V$ graph is the reverse of the lamp's: as the current heats it, its resistance falls, so the curve bends towards the $I$ -axis. This follows from the behaviour of thermistors explained in Resistivity, thermistors and LDRs.

Finding resistance from an I–V graph

At any point on the graph, $R = V/I$ : read off the coordinates and divide.

Watch out

The resistance at a point on a curve is $V/I$ at that point, not the inverse of the gradient of the tangent there. The two agree only for a straight line through the origin. In the lamp example below, at $10\ \text{V}$ the resistance is $30\ \Omega$ , but $1/\text{gradient}$ of the tangent is $90\ \Omega$ . Examiners report this error every year.

On an $I$ – $V$ graph ( $I$ up, $V$ across), the gradient of the line joining the origin to a point is $I/V = 1/R$ . So "the line from the origin gets less steep" means "the resistance increases". If the axes are swapped ( $V$ up, $I$ across), the gradient of that line is $R$ directly. Always check which way round the axes are.

Using V = IR

(a) A resistor carries a current of $0.48\ \text{A}$ when the p.d. across it is $12\ \text{V}$ . Calculate its resistance. (b) A $47\ \Omega$ resistor carries a current of $250\ \text{mA}$ . Calculate the p.d. across it.

Solution

(a)

R = \frac{V}{I} = \frac{12}{0.48} = 25\ \Omega

(b) Convert the current: $250\ \text{mA} = 0.250\ \text{A}$ .

V = IR = 0.250 \times 47 = 11.75\ \text{V} \approx 12\ \text{V}

Resistance of a filament lamp

Readings for a filament lamp are shown.

$V$ / V	0	2.0	4.0	6.0	8.0	10.0	12.0
$I$ / A	0	0.143	0.222	0.273	0.308	0.333	0.353

(a) Calculate the resistance of the lamp at $2.0\ \text{V}$ and at $10.0\ \text{V}$ . (b) Explain the change. (c) The gradient of the tangent to the $I$ – $V$ curve at $10.0\ \text{V}$ is $0.0111\ \text{A V}^{-1}$ . A student says the resistance at $10.0\ \text{V}$ is $1/0.0111 = 90\ \Omega$ . Explain the error.

Solution

(a)

R_{2} = \frac{2.0}{0.143} = 14\ \Omega, \qquad R_{10} = \frac{10.0}{0.333} = 30\ \Omega

(b) The larger current at $10.0\ \text{V}$ heats the filament to a higher temperature. The lattice ions vibrate with larger amplitude, the electrons collide with them more frequently, and the resistance increases.

(c) Resistance is defined as $V/I$ , the ratio of the coordinates of the point, not the reciprocal of the gradient of the tangent. The tangent gradient describes how the current changes with p.d., which is a different quantity for a curved graph. The resistance at $10.0\ \text{V}$ is $30\ \Omega$ .

A diode with a series resistor

A silicon diode is connected in series with a $220\ \Omega$ resistor and a $5.0\ \text{V}$ battery so that the diode is forward biased. The p.d. across the conducting diode is $0.70\ \text{V}$ . (a) Calculate the current. (b) The battery is reversed. State the current and the p.d. across the diode.

Solution

(a) The p.d. across the resistor is the rest of the battery p.d.: $5.0 - 0.70 = 4.3\ \text{V}$ .

I = \frac{4.3}{220} = 0.0195\ \text{A} = 20\ \text{mA}

(b) Reverse biased, the diode has a very high resistance: the current is (almost) zero. With no current, the p.d. across the resistor is $IR = 0$ , so the whole $5.0\ \text{V}$ appears across the diode.

A lamp in series with a resistor

The lamp from the earlier example is connected in series with a $20\ \Omega$ resistor across a $12.0\ \text{V}$ supply of negligible internal resistance. Use its characteristic to find the current in the circuit and the p.d. across the lamp.

Solution

The lamp is non-ohmic, so you cannot simply add resistances: its resistance depends on the current, which is what you are trying to find. Use the graph.

The current $I$ is the same in the lamp and the resistor, and the p.d.s must add to $12.0\ \text{V}$ :

V_{\text{lamp}} + 20I = 12.0 \qquad \Rightarrow \qquad I = \frac{12.0 - V_{\text{lamp}}}{20}

This is a straight line on the $I$ – $V$ graph from $(0,\ 0.60)$ to $(12.0,\ 0)$ . The operating point is where it crosses the lamp's characteristic.

y = x / (10 + 2 x) y = (12 - x) / 20 (6.39, 0) -- (6.39, 0.2805)

The intersection is at $V_{\text{lamp}} \approx 6.4\ \text{V}$ , $I \approx 0.28\ \text{A}$ .

Check: p.d. across the resistor $= 0.28 \times 20 = 5.6\ \text{V}$ , and $6.4 + 5.6 = 12.0\ \text{V}$ .

Without a graph you can get the same result by trial using the table: at $V = 6.0\ \text{V}$ , $V + 20I = 6.0 + 20 \times 0.273 = 11.5\ \text{V}$ (too small); at $8.0\ \text{V}$ , $8.0 + 20 \times 0.308 = 14.2\ \text{V}$ (too big); so the answer lies a little above $6\ \text{V}$ .

Exam tip

Sketch questions: label both axes ( $I$ and $V$ ), pass through the origin, and show the correct shape in both quadrants for the resistor and lamp (the lamp is symmetric). For the diode, zero current in reverse and a sharp rise after about $0.6\ \text{V}$ in forward bias. A sketch does not need a scale but must have the key features.
"Explain why the resistance of a filament lamp increases" earns marks for: current increases, temperature increases, (lattice) ions vibrate more, more frequent collisions with electrons, so resistance increases. Say "as the current increases" or "as the p.d. increases"; the syllabus phrasing links it to current.
"State Ohm's law": include proportional and constant temperature (or physical conditions). Missing the condition loses a mark.
Before calculating a resistance from a graph, write $R = V/I$ and quote the coordinates you used.

Summary

Resistance is the ratio of p.d. to current, $R = V/I$ ; $1\ \Omega = 1\ \text{V A}^{-1}$ . $V = IR$ applies to every component.
Ohm's law: $I \propto V$ for a conductor provided physical conditions such as temperature stay constant.
Metal at constant temperature: straight line through the origin (ohmic).
Filament lamp: S-shaped curve; resistance increases with current because the temperature rises and electrons collide more often with the vibrating ions.
Semiconductor diode: no current in reverse; conducts strongly above about $0.6$ to $0.7\ \text{V}$ forward.
On a curve, $R = V/I$ at the point, never $1/\text{gradient}$ of the tangent.
A non-ohmic component in a circuit is analysed by using its characteristic, not a fixed resistance.

Practice

Question

Define resistance and state its unit in SI base units.
A heater has a resistance of $26\ \Omega$ and is connected to $230\ \text{V}$ . Calculate the current.
State Ohm's law, and explain why a filament lamp does not obey it.
Sketch, on the same axes, the $I$ – $V$ characteristics of a $10\ \Omega$ resistor and a filament lamp that has a resistance of $10\ \Omega$ at $3.0\ \text{V}$ . Mark where the curves cross.
Using the lamp table in the worked examples, find the resistance of the lamp at $6.0\ \text{V}$ and the power it dissipates.
A diode is in series with a resistor across a $3.0\ \text{V}$ cell, forward biased. The current is $15\ \text{mA}$ and the diode p.d. is $0.65\ \text{V}$ . Calculate the resistance of the resistor.
A student plots $V$ on the $y$ -axis against $I$ on the $x$ -axis for a filament lamp. Describe the shape of the graph and explain how the resistance at a point is found from it.
Two diodes D1 and D2 and a lamp are connected in a circuit with a battery. D1 is in series with the lamp, pointing so it is forward biased. D2 is connected in parallel with the lamp, also pointing in the direction of conventional current from the positive terminal. Describe what happens to the lamp, and explain what happens if the battery is reversed.
The lamp in the worked examples has resistance $R = (10 + 2V)\ \Omega$ for $V \ge 0$ . It is connected in parallel with a $30\ \Omega$ resistor to a variable supply. (a) At what p.d. do the lamp and the resistor carry equal currents? (b) At $4.0\ \text{V}$ , calculate the total current drawn from the supply.
The lamp from the worked examples is connected in series with an identical lamp across a $12.0\ \text{V}$ supply. A student argues that each lamp has $6.0\ \text{V}$ across it, so the current is $0.273\ \text{A}$ . (a) Explain why the student is right in this case. (b) The second lamp is replaced by a $30\ \Omega$ resistor. Use the formula $I = V/(10 + 2V)$ to find the p.d. across the lamp, and comment on how this compares with the lamp in series with a $20\ \Omega$ resistor.

Answers

Resistance is the ratio of the p.d. across a component to the current in it, $R = V/I$ . Unit: $\Omega = \text{V A}^{-1} = \text{kg m}^2\,\text{s}^{-3}\,\text{A}^{-2}$ .
$I = V/R = 230/26 = 8.8\ \text{A}$ .
The current in a conductor is directly proportional to the p.d. across it provided physical conditions such as temperature remain constant. In a lamp, increasing the current raises the filament's temperature, so its resistance increases and $I$ is not proportional to $V$ .
The resistor is a straight line through the origin with gradient $0.1\ \text{A V}^{-1}$ . The lamp curve starts steeper than the line (lower resistance when cool), crosses it at $(3.0\ \text{V},\ 0.30\ \text{A})$ , and lies below it at higher p.d.s (higher resistance when hot). Both are symmetric through the origin, so they also cross at $(-3.0\ \text{V},\ -0.30\ \text{A})$ .
$R = 6.0/0.273 = 22\ \Omega$ ; $P = VI = 6.0 \times 0.273 = 1.6\ \text{W}$ .
P.d. across resistor $= 3.0 - 0.65 = 2.35\ \text{V}$ ; $R = 2.35/0.015 = 157 \approx 160\ \Omega$ .
The graph starts as a straight line from the origin and then curves upwards (towards the $V$ -axis), becoming steeper, because the resistance increases. The resistance at a point is $V/I$ from the coordinates, which equals the gradient of the line from the origin to that point (not the gradient of the tangent).
D2 is forward biased and in parallel with the lamp; its p.d. cannot rise much above about $0.7\ \text{V}$ , and it offers a low-resistance path, so almost all the current passes through D2: the lamp is very dim or does not light. (D1 conducts.) With the battery reversed, D1 is reverse biased and in series with everything, so no current flows anywhere and the lamp is off.
(a) Equal currents means equal resistances (same $V$ ): $10 + 2V = 30$ , so $V = 10\ \text{V}$ . (b) Lamp: $R = 18\ \Omega$ , $I = 4.0/18 = 0.222\ \text{A}$ . Resistor: $I = 4.0/30 = 0.133\ \text{A}$ . Total $= 0.356\ \text{A} \approx 0.36\ \text{A}$ .
(a) The lamps are identical and carry the same current, so they must have identical p.d.s; each has $6.0\ \text{V}$ and the table gives $0.273\ \text{A}$ . (b) $V + 30\dfrac{V}{10 + 2V} = 12$ gives $V(10 + 2V) + 30V = 12(10 + 2V)$ , so $2V^2 + 16V - 120 = 0$ , i.e. $V^2 + 8V - 60 = 0$ , so $V = \dfrac{-8 + \sqrt{64 + 240}}{2} = 4.72\ \text{V}$ . Current $= 4.72/19.4 = 0.243\ \text{A}$ ; resistor p.d. $= 7.28\ \text{V}$ (check: total $12.0\ \text{V}$ ). With the larger series resistor, the lamp gets a smaller share of the p.d. ( $4.7\ \text{V}$ rather than $6.4\ \text{V}$ ) and a smaller current ( $0.24\ \text{A}$ rather than $0.28\ \text{A}$ ).

Resistance

What causes resistance

Ohm's law

Measuring an I–V characteristic

Three characteristics to know

Metallic conductor at constant temperature

Filament lamp

Semiconductor diode

Finding resistance from an I–V graph

Summary

Practice

Console