Current, resistance and potential difference: The current (I) through a component depends on both the resistance (R) of the component and the potential difference (V) across the component. The greater the resistance of the component the smaller the current for a given potential difference (pd) across the component.

The idea

The current through a component is set by two things together: the potential difference across it and its resistance. Potential difference provides the push that drives charge, while resistance opposes the flow. For a fixed pd, increasing resistance reduces current; for a fixed resistance, increasing pd increases current.

What earns the marks

• State that current through a component depends on both the potential difference across it and its resistance.
• Use the equation V = I × R and its rearrangements to calculate current, potential difference or resistance.
• Explain that for a given potential difference, a greater resistance produces a smaller current.
• Describe resistance as opposition to the flow of charge, measured in ohms (Ω).
• Apply the relationship to a numerical example, such as 12 V across 4 Ω giving 3 A, and predict the effect of increasing resistance.

Common mistakes and how to put them right

• Saying that increasing resistance increases current.

  Put it right by explaining that resistance opposes charge flow, so current falls for a fixed pd.

• Mixing up the symbols V, I and R in the equation.

  Put it right by writing V = I × R and checking units — volts, amperes and ohms.

• Assuming current is directly proportional to resistance.

  Put it right by noting that current is inversely proportional to resistance when pd is constant.

Exam tips

• Write the equation, substitute values with units, then rearrange only after substitution to reduce errors.
• When explaining a change, state which quantity is held constant before describing the effect on current.
• Check that your calculated current is smaller when resistance is larger for the same pd, as a quick sanity check.