A-level Physics (Advancing Physics)/Resistance and Conductance

Conductance is a measure of how well an artefact (such as an electrical component, not a material, such as iron) carries an electric current. Resistance is a measure of how well an artefact resists an electric current.

Resistance is measured in Ohms (usually abbreviated using the Greek letter Omega, Ω) and, in formulae, is represented by the letter R. Conductance is measured in Siemens (usually abbreviated S) and, in formulae, is represented by the letter G.

Resistance and conductance are each other's reciprocals, so:

$R={\frac {1}{G}}$ and $G={\frac {1}{R}}$

Ohm's Law

Ohm's Law states that the potential difference across an artefact constructed from Ohmic conductors (i.e. conductors that obey Ohm's Law) is equal to the product of the current running through the component and the resistance of the component. As a formula:

V = IR

where V is potential difference (in V), I is current (in A) and R is resistance (in Ω).

In terms of Resistance

This formula can be rearranged to give a formula which can be used to calculate the resistance of an artefact:

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

In terms of Conductance

Since conductance is the reciprocal of resistance, we can deduce a formula for conductance (G):

${\frac {1}{G}}={\frac {V}{I}}$

$G={\frac {I}{V}}$

The Relationship between Potential Difference and Current

From Ohm's Law, we can see that potential difference is directly proportional to current, provided resistance is constant. This is because two variables (let us call them x and y) are considered directly proportional to one another if:

$y=kx$

where k is any positive constant. Since we are assuming that resistance is constant, R can equal k, so V=RI states that potential difference is directly proportional to current. As a result, if potential difference is plotted against current on a graph, it will give a straight line with a positive gradient which passes through the origin. The gradient will equal the resistance.

In Series Circuits

In a series circuit (for example, a row of resistors connected to each other), the resistances of the resistors add up to give the total resistance. Since conductance is the reciprocal of resistance, the reciprocals of the conductances add up to give the reciprocal of the total conductance. So:

$\Sigma R=R_{1}+R_{2}+...+R_{n}$

$\Sigma {\frac {1}{G}}={\frac {1}{G_{1}}}+{\frac {1}{G_{2}}}+...+{\frac {1}{G_{n}}}$

In Parallel Circuits

In a parallel circuit, the conductances of the components on each branch add up to give the total conductance. Similar to series circuits, the reciprocals of the total resistances of each branch add up to give the reciprocal of the total resistance of the circuit. So:

$\Sigma G=G_{1}+G_{2}+...+G_{n}$

$\Sigma {\frac {1}{R}}={\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}+...+{\frac {1}{R_{n}}}$

When considering circuits which are a combination of series and parallel circuits, consider each branch as a separate component, and work out its total resistance or conductance before finishing the process as normal.

Questions

1. The potential difference across a resistor is 4V, and the current is 10A. What is the resistance of the resistor?

2. What is the conductance of this resistor?

3. A conductor has a conductance of 2S, and the potential difference across it is 0.5V. How much current is flowing through it?

4. A graph is drawn of potential difference across an Ohmic conductor, and current. For every 3 cm across, the graph rises by 2 cm. What is the conductance of the conductor?

5. On another graph of potential difference and current, the graph curves so that the gradient increases as current increases. What can you say about the resistor?

6. 3 resistors, wired in series, have resistances of 1kΩ, 5kΩ and 500Ω each. What is the total resistance across all three resistors?

7. 2 conductors, wired in parallel, have conductances of 10S and 5S. What is the total resistance of both branches of the parallel circuit?

8. The circuit above is attached in series to 1 10Ω resistor. What is the total conductance of the circuit now?

Worked Solutions