6.1.1 Capacitors

Capacitors are electrical components which separate charge. They consist of two metallic plates separated by an insulator (a dielectric - e.g air/ceramic/paper/mica). When a capacitor is connected to a cell of e.m.f Є (for example) the electrons only flow from the cell for a short time as they cannot travel between plates (because of the insulator). During the brief current, electrons from the cell flow onto one plate (they are deposited onto a plate, this plate has a net negative charge as it gains electrons) and this repels electrons from the other plate so electrons are removed from the other plate (this plate has a net positive charge as it is deficient in electrons). Charge is always conserved - we can confirm this as the current in the circuit must be the same at all points. The same amount of electrons that has been deposited onto one plate has left the other plate so the have an equal but opposite charge (they have a net charge of 0). This means there is a p.d. across the plates. When the p.d. across the plates is equal to the e.m.f (Є) of the cell, the current in the circuit falls to 0 and the capacitor is fully charged.

The capacitance of a capacitor is defined as the charge stored per unit p.d. across it. It is measured in farads (F). From the equation below we can see that one Farad equates to one coulomb per volt. We can use the following equation to determine capacitance:

Q = V C

The greater the amount of positive and negative charge stored on the plates the greater the p.d. across them. This means that charge is proportional to p.d meaning that, from the equation above, capacitance is constant.

Okay so when we are connecting capacitors in circuits we cannot get them mixed up with resistors!! This is a big no no, they have the opposite rules...

• In parallel the total capacitance is C = C1 + C2.....
• This is because the pd. across each capacitor is the same and charge is conserved so total Q (total charge stored) equates to the sum of the individual charges stored by each capacitor....Q = Q1 + Q2... Therefore since Q = VC and V is constant then C = C1 + C2....
• In series the total capacitance is 1/C = 1/C1 + 1/C2.......
• This is because in series the sum of the p.d.s around each loop equals the e.m.f so V = V1 + V2.... We know that the charge stored in each capacitor is the same so Q is constant. Q = VC so V = Q/C so 1/C = 1/C1 + 1/C2...

There's a little bit more about capacitors in circuits. Firstly, how to investigate combinations/perhaps an unknown capacitor value. Set up a circuit with a safety resistor, an ammeter, a few capacitors and a voltmeter across each capacitor and a variable power supply (a power supply where we can vary the output voltage) and also a switch. Close the switch - current will briefly flow through the circuit. We can determine the charge stored in each capacitor by measuring the voltage across it and multiply it by the capacitor reading (Q = V C). You will see that in each instance in SERIES the charge stored in each capacitor will be the same for a certain voltage (this value will vary as voltage across the circuit varies).

To determine the series rule for capacitors (1/C = 1/C1 + 1/C2...) connect a multimeter set to capacitance across two capacitors in series. The reading will show the figure obtained is we were to do the sum 1/C = 1/C1 + 1/C2...

6.1.3 Charging and discharging capacitors

Discharging capacitors
Capacitors discharge through a constant-ratio pattern known as exponential decay.

Before a capacitor is allowed to discharge through a resistor (e.g if a switch is open at t=0)...

• The p.d. (V) across the capacitor or resistor is V₀
• For the capacitor Q = V C so charge stored is the capacitor is Q = V₀C
• For the resistor V=IR so the current in the resistor is I = V₀/R

When a switch is closed the capacitor discharges through the resistor. This means that, as time goes by, the charge stored by the capacitor decreases (so the p.d. across it also decreases). Since the p.d. decreases, the current in the resistor decreases accordingly and eventually p.d., the charge stored in the capacitor, and the current in the resistor are all 0. They all show exponential decay over time meaning we can use similar equations for each quantity...

V = V₀e^-t/CR

I = I₀e^-t/CR

Q = Q₀e^-t/CR 

Where I₀, V₀, and Q₀ is the maximum current/p.d./charge at t=0.

The time constant of a capacitor-resistor circuit is CR (in seconds). It is a measure of how long the exponential decay will take in a particular capacitor-resistor circuit. It's symbol is τ (tau) and it is measured in seconds (s).When t = CR, the p.d. across the resistor/capacitor is given by...

V = V₀e^-t/CR = V₀e^-CR/CR = V₀e^-1 = 0.37 V₀

Okay so we need to be able to model exponential decay. Provided the capacitor and resistor in a circuit are in parallel then they will have the same p.d. We known that Q = V C (charged stored by a capacitor) and V = I R (current stored in a circuit). Since V is the same for both components we can form the following equation for current:

I = V/R = Q/(CR)

Usually, I = ΔQ/Δt. However, for a capacitor, I = -ΔQ/Δt (this shows that the charge on the capacitor decreases with time). We can now write the above equation as:

-ΔQ/Δt = Q/(CR)

Meaning that...

ΔQ/Δt = -Q/(CR)

Q = Q₀e^-t/CR is simply a solution to this equation.

This equation (ΔQ/Δt = -Q/(CR)) can be used to model the decay of Q (charge on a capacitor)...

• Start with a known value Q₀ of for the time constant CR
• choose a time interval Δt which is small compared to CR
• Calculate the charge (ΔQ) leaving the capacitor in the time interval (Δt)...
• ΔQ = (Δt/(CR)) x Q
• Calculate Q left on capacitor by subtracting ΔQ from the previous charge
• Repeat many times

Straight line graphs

We know that the p.d. (V) across a discharging capacitor is V = V₀e^-t/CR. If we take logs  (to the base e) of both sides we get...

lnV = ln(V₀e^-t/CR) = lnV₀ + ln e^-t/CR = lnV₀ - (t/(CR))

We can plot ln V against t and the gradient will be -1/(CR) and the y-intercept will be lnV₀.

Charging capacitors

Okay so all that stuff above is all well and good if we want to talk about discharging capacitors, but what about actually charging them in the first place? Well, when we close a switch in a circuit the capacitor will start to charge. The p.d. across the capacitor (V_C)will increase from 0. The p.d. across the capacitor (V_C) and resistor (V_R) must always add up to V₀ (K2) so it follows that V_R must decrease as V_C increases.  After a while V_R falls to 0 as the capacitor is fully charged (V₀ = V_C).

We know that in a circuit current decreases exponentially as the capacitor discharges through the resistor...I = I₀e^-t/CR. Since V = IR and R is constant we know that...

V_R = V₀e^-t/CR

At any time (t) V₀ = V_R + V_C  so...

V_C = V₀ - V₀e^-t/CR

Therefore...

V_C = V₀ (1 - e^-t/CR)

We can also use this equation with charge Q on a capacitor (as well as p.d. V across a capacitor).

6.1.2 Energy

Okay so this whole section is basically about how to use graphs with capacitors...then a little bit on what capacitors are useful for.

The amount of energy stored in a capacitor depends on the value of the capacitance and the initial p.d. across it.

When an electron moves from an emf source (e.g a power pack/battery) on to the capacitor plate, it will experience a repulsive force from the electrons that have already been deposited onto the plate since they have the same charge. This means that work has to be done to push the electron onto the plate. This work is supplied by the battery. Basically the energy stored in a capacitor comes from the battery/power supply.

We can determine the energy stored in a capacitor using a graph of p.d. against charge (for a capacitor, duh). This gives a triangular like shape and the work done is the work under this graph. The work done on the charges is the same as the energy stored in the capacitor (this is potential/stored energy)...

area under graph = area of shaded triangle = W = 0.5VQ

If we use Q = V C we can work out a few more variations of this equation...

W =  0.5Q^2/C = 0.5 V^2C

Uses of capacitors
So it's all very good that we know all about capacitors, but why are they actually useful? Well, capacitors release the stored energy very quickly which generates a high output power. This is useful in (for example)...

• camera flashes
• back up power for computers
• emergency lighting if the mains supply cuts out

Okay so the spec technically doesn't say we need to know about smoothing capacitors but I think it's useful to know so here we go...

Mains electricity is AC (supply voltage changes rapidly from positive to negative changing the direction of current). If there is a diode in the circuit it will only allow current flowing in one direction. Without a capacitor output voltage will consist of positive cycles only - with a capacitor the output voltage is smoothed out and is almost completely DC (direct current) with a constant value. By making the time constant of the circuit much greater than the period of the alternating voltage the 'ripple' in output voltage can be kept small. It is important that the time constant be much greater than the period of the input alternating voltage else the output voltage would not be very smooth.

NOTE: the 'ripple' is the difference between the maximum and minimum output voltage