Capacitors in series and parallel, energy stored, and RC decay

Derivation: capacitors in parallel

C₁, C₂ and C₃ are joined between the same two points, across a supply V.

1. 1
   Each capacitor has the same potential difference V.
2. 2
   The charges are Q₁ = C₁V, Q₂ = C₂V and Q₃ = C₃V.
3. 3
   Total charge: Q = Q₁ + Q₂ + Q₃ = (C₁ + C₂ + C₃)V.
4. 4
   For the equivalent capacitor, Q = CV.
5. 5★
   So, C = C₁ + C₂ + C₃. The equivalent is larger than the largest capacitor.

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Derivation: capacitors in series

C₁, C₂ and C₃ are joined end to end across a supply of potential difference V.

1. 1
   Each capacitor gets the same charge Q: the joined inner plates of two neighbouring capacitors are isolated, so their charges (−Q and +Q) add up to zero.
2. 2
   The potential differences are V₁ = Q/C₁, V₂ = Q/C₂ and V₃ = Q/C₃.
3. 3
   They add up to the supply: V = V₁ + V₂ + V₃ = Q(1/C₁ + 1/C₂ + 1/C₃).
4. 4
   For the equivalent capacitor C, V = Q/C.
5. 5★
   So, 1/C = 1/C₁ + 1/C₂ + 1/C₃. The equivalent is smaller than the smallest capacitor.

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How to draw series and parallel combinations

• Capacitor symbol: two short parallel lines of equal length, with a gap between them.
• Series: draw C₁, C₂ and C₃ one after another on a single wire, with the battery across the ends. Mark +Q and −Q on each, and V₁, V₂, V₃ below.
• Parallel: draw two horizontal wires joined to the battery, and put C₁, C₂ and C₃ as vertical rungs between them. Mark Q₁, Q₂, Q₃ and the common V.

Derivation: energy stored in a capacitor

An uncharged capacitor C is joined to a battery and slowly charged to a charge q and a potential difference V.

1. 1
   The battery moves electrons from one plate to the other. It must do work, because the charge already there repels the new charge.
2. 2
   During charging, the potential difference rises steadily from 0 to V.
3. 3
   So the average potential difference is (0 + V)/2 = V/2 = q/(2C).
4. 4
   Work done = charge × average potential difference = q × q/(2C) = q²/(2C).
5. 5★
   So, the energy stored is U = q²/(2C) = ½qV = ½CV². It sits in the electric field between the plates.
6. 6
   Touching the plates of a charged, high-value capacitor is dangerous: it can discharge through your body.

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Worked example: three capacitors, charge and energy

Activity: decay of current in an RC circuit

Aim: to watch the charging current of a capacitor die away and find the time constant. You need a 2000 μF, 10 V electrolytic capacitor, a high-resistance box, a galvanometer, a 3 V battery, two keys and a stop watch.

1. 1
   Join the battery, a main key, the resistance box R, the galvanometer and the capacitor in series. Join the capacitor's + terminal towards the battery's + terminal.
2. 2
   Join a second key across the capacitor and close it, so the capacitor stays uncharged. Take out about 10 000 Ω from the box to protect the galvanometer.
3. 3
   Close the main key. Adjust R to get a full-scale deflection, say 30 divisions.
4. 4
   Open the key across the capacitor and start the stop watch at the same moment. The capacitor begins to charge.
5. 5
   Record the time when the deflection falls to 24, 20, 16, 12, 10, 8, 6, 4, 3 and 2 divisions. Repeat twice and take the mean times.
6. 6
   Plot deflection θ on the y-axis against time t on the x-axis. Observe: the curve falls quickly at first, then more and more slowly.
7. 7
   From the graph, read the time for the deflection to fall to half of any starting value (the half-life), and to 0.37 of it (the time constant).
8. 8★
   Result: the current falls as I = I₀ exp(−t/RC). The time constant comes out close to RC, and the half-life close to 0.693RC. For R = 10 000 Ω and C = 2000 μF, RC = 20 s.

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