# Solution to #1-5

Paper 2 Difficulty: Medium

The current passing through a resistor is $I=3\pm0.1\ \textrm A$ and the resistance of the resistor is $R=13\pm0.5\ \Omega$. The electrical power, measured in watt (W), supplied to the resistor is given by $P=RI^{2}$.

a) Write down the value of the supplied power correct to one significant figure.

[Maximum mark: 2]

$P=RI^{2}\\ P=13\times 3^{2}\\ P=117\ \textrm W$

b) Find the percentage uncertainty for the current passing through the resistor and its resistance.

[Maximum mark: 4]

$\frac{\Delta I}{I}=\frac{0.1}{3}\approx0.0333\\ 0.0333\times 100=3.33\%\\ \frac{\Delta R}{R}=\frac{0.5}{13}\approx0.0385\\ 0.0385\times 100=3.85\%$

c) Find the absolute uncertainty for the electrical power.

[Maximum mark: 2]

$\frac{\Delta P}{P}=\frac{\Delta R}{R}+2\times \frac{\Delta I}{I}\\ \frac{\Delta P}{P}= 0.0385+2\times0.0333=0.1051\\ \Delta P=\frac{\Delta P}{P}\times P=0.1051\times117\\\Delta P\approx12.3\ \textrm W$