Capacitors Tutorial 4 - Capacitor Circuits

Please read the extension material for a more in depth treatment of the topic.

This is on the OCR Syllabus and the Welsh Syllabus.

 Contents Parallel Capacitors Series Capacitors Deriving the Equations

### Parallel Capacitors

Here is a circuit consisting of two capacitors in parallel. They have values C1 and C2 and are connected to a battery of voltage V. • Like all parallel circuits the voltage across the capacitors is the same.

• The total charge is the sum of the charges on the capacitors. It’s like the currents in parallel resistors adding up.

Ctot = C1 + C2

This is true for any number of parallel capacitors, so

Ctot = C1 + C2 + C3 + … + Cn

 Worked Example What is the single capacitor equivalent of this circuit below? What is the charge on each capacitor? Answer Use Ctot = C1 + C2 to get the capacitance Use Q = CV to get the charge.                  Ctot = C1 + C2 = 4 mF + 6 mF = 10 mF             Charge on C1 = 4 × 10-6 F × 12 V = 4.8 × 10-5 C = 48 mC             Charge on C2 = 6 × 10-6 F × 12 V = 7.2 × 10-5 C = 72 mC             Total charge = 48 mC + 72 mC = 120 mC

 Question 1 What is the single capacitor equivalent of these parallel capacitors? ### Series Capacitors

Here is a circuit consisting of two capacitors in series. They have values C1 and C2 and are connected to a battery of voltage V. In any series circuit

• The voltages add up to the battery voltage;

• The current (charge) is the same all the way round.

Since Q = It, it is reasonable to say that the charge that has moved is the same all the way round. If a number of electrons of total charge of Q crowds onto the negative plates of C2 then the same number of electrons are repelled away from the positive plates. These crowd onto the negative plates of C1 and repel the same number away from the positive plates.

Now we know that:

V = Q/C

And that:

Vtot = V1 + V2.

So we can write:

This gives us a general relationship for any number of series capacitors:

We can tackle problems that involve both series and parallel capacitors in a similar way to the way we tackle problems with combined series and parallel arrays of resistors.

 Worked Example What is the single capacitor equivalent of this circuit below? What is the charge on each capacitor? What are the voltmeter readings? Answer Use 1/Ctot = 1/C1 + 1/C2 Voltages add up to the battery voltage.   Work out the total capacitance: Þ Ctot = 12/5 = 2.4 mF           Now we can work out the charge:   Q = CV = 2.4 mF × 12 V = 28.8 mC   Now work out the voltages: On capacitor 1, V = 28.8 mC ÷ 4 mF = 7.2 V On capacitor 2, V = 28.8 mC ÷ 6 mF = 4.8 V   Voltages add up. Battery voltage = 7.2 + 4.8 = 12.0 V (Kirchhoff II)

 Question 2 What is the single capacitor equivalent of these series capacitors? When you tackle problems involving both series and parallel capacitors in the same circuit, you may find it helpful to adopt the following problem solving strategy:

• Work out the single capacitor equivalent of the parallel capacitors

• Then use this answer to work out the single capacitor equivalent of the series capacitors.