In a laboratory experiment to verify Ohm's law, a student measures...

Given $V = 90 \pm 1.8 \text{ V}$ and $I = 3.0 \pm 0.15 \text{ A}$.

From Ohm's law, the resistance is given by $R = \dfrac{V}{I}$.

Substituting the nominal values:

$R = \dfrac{90}{3.0} = 30 \Omega$

The maximum fractional error in resistance is the sum of the fractional errors in voltage and current:

$\dfrac{\Delta R}{R} = \dfrac{\Delta V}{V} + \dfrac{\Delta I}{I}$

$\dfrac{\Delta R}{30} = \dfrac{1.8}{90} + \dfrac{0.15}{3.0}$

$\dfrac{\Delta R}{30} = 0.02 + 0.05 = 0.07$

The absolute error in resistance is:

$\Delta R = 30 \times 0.07 = 2.1 \Omega$

Thus, the calculated resistance with limits of error is $(30 \pm 2.1) \Omega$.

Answer: $(30 \pm 2.1) \Omega$