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Physics › Capacitance › Charging and discharging

Charging and discharging

Put a resistor in the way and a capacitor fills and empties along curves, never lines. Three quantities tell the story, and the exam's favourite trick is the one that moves opposite to the other two.

Year 13AQA 3.7.4.4

Builds on Capacitors and energy stored and Circuits and Kirchhoff's laws.

IN THIS TOPIC

  • Sketch and explain the Q, V and I against t graphs for discharge.
  • Sketch and explain the charging graphs, including the current's opposite behaviour.
  • Read gradients and areas: current from the Q–t gradient, charge from the I–t area.

WHAT YOU PROBABLY THINK

A capacitor charges at a steady rate until it is full.

Discharge: one shape, three names

Discharging through a resistor: charge, pd and current all fall along the same exponential curvetQ, V and I all share this one shapesteep while full, gentle while empty
FIG. 1Discharge through a resistor: one exponential decay, shared by the charge, the pd and the current.

Let a charged capacitor drive current through a resistor and three quantities fall together. The pd tracks the charge through V = Q/C; the current tracks the pd through I = V/R; so Q, V and I all follow the same falling curve. The shape explains itself: a full capacitor pushes a large current and empties quickly, but the emptier it gets, the smaller the pd, the smaller the current, and the slower the emptying. Each step of loss weakens the very thing driving the loss, and that self-throttling is what makes the decay exponential: steep while full, gentle while nearly empty, never quite reaching zero.

Charging: the odd one out

Charging through a resistor: charge and pd climb toward their final values while the current starts at its maximum and dies awaytfinal valueQ and V climbI dies away
FIG. 2Charging through a resistor: charge and pd climb toward their final values while the current starts at its maximum and dies.

Connect an uncharged capacitor to a supply through a resistor and the charge and the capacitor pd climb toward their final values, mirroring the discharge curves. The current does the opposite: at the first instant the capacitor is uncharged, offers no opposing pd, and the current starts at its maximum, I0 = V/R. As charge builds, the capacitor's growing pd opposes the supply, the difference across the resistor shrinks, and the current dies away toward zero. Fully charged means the capacitor pd equals the supply and nothing flows. Examiners love the asymmetry: two quantities rise, one falls, and the lie above fails on sight of any of the three, since nothing here happens at a steady rate.

Gradients and areas

The spec asks you to read these graphs, not only sketch them, and the dictionary is the one you already own. On the Q–t graph the gradient is the current, since I = ΔQ/Δt: steepest at the start of either process, flattening to zero. On the I–t graph the area underneath is the charge transferred, the same reading as in the electricity unit. During a full charge, the whole area under the decaying current curve equals the final charge CV, however the current wriggles on the way.

THE EXAM BIT

  • Sketching marks come from the landmarks: correct starting value, correct final value, steepest gradient at t = 0, and a curve that flattens without touching its asymptote.
  • State the direction of every curve before its shape: discharge sends Q, V and I down together; charging sends Q and V up while I falls.
  • The explanation mark for the shape is the feedback sentence: the current depends on the pd, the pd depends on the charge, so the rate of change falls as the process runs.
  • Charging current at t = 0 is I0 = V/R, set by the resistor alone, because an uncharged capacitor opposes nothing at that instant.
  • Gradient of Q–t gives I; area under I–t gives Q. Quote whichever direction the data supports, exactly as with motion graphs.

CHECK YOURSELF

A 12 V supply charges a capacitor through a 10 kΩ resistor. State the current the moment the switch closes, and the current and capacitor pd after a long time, explaining each value.

Show a hint

At the first instant the capacitor opposes nothing; after a long time it opposes everything.

Show the answer

At t = 0 the uncharged capacitor has no pd, so the whole 12 V sits across the resistor: I0 = V/R = 12 / 10 000 = 1.2 mA.

After a long time the capacitor pd has climbed to 12 V, matching the supply.

With no pd left across the resistor the current is zero: the capacitor is fully charged and the circuit rests.

Discharge: Q, V and I fall along one shared curve.

Charging: Q and V climb while the current starts big and dies.

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