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You connect a mystery resistor to a power supply, crank the voltage, and plot the readings. How can the angle of a single straight line immediately tell you the exact resistance of the component?
In 1827, German physicist Georg Ohm discovered that for a metal wire kept at a constant temperature, the current flowing through it is directly proportional to the potential difference across it.
This relationship is Ohm's Law, written as V=IΓ—RV = I \times R, where VV is potential difference in volts (V), II is current in amperes (A), and RR is resistance in ohms (Ξ©\Omega).
Physics exams often swap which variable sits on the vertical axis, completely changing how you calculate resistance. How do you handle a graph when voltage is on the vertical axis?

Reading V-I Graphs Directly

On a V-I graph, voltage VV is plotted on the vertical yy-axis and current II is plotted on the horizontal xx-axis.
The gradient (or slope) of a straight line is the change in the vertical value divided by the change in the horizontal value: gradient=Ξ”yΞ”x\text{gradient} = \frac{\Delta y}{\Delta x}.
Because gradient=Ξ”VΞ”I\text{gradient} = \frac{\Delta V}{\Delta I} and Ohm's Law states R=VIR = \frac{V}{I}, the gradient of a V-IV\text{-}I graph gives resistance directly: R=gradientR = \text{gradient}.
For the line shown in the diagram, the resistance calculation is simply R=6.0Β Vβˆ’0Β V0.5Β Aβˆ’0Β A=12Β Ξ©R = \frac{6.0\text{ V} - 0\text{ V}}{0.5\text{ A} - 0\text{ A}} = 12\ \Omega.
What happens when current is placed on the vertical axis instead?

Inverting the Gradient on I-V Graphs