This circuit can be used to measure the value of a capacitor:
The reed switch is operated from a 400 Hz supply.
It operates on the forward half cycle, to charge up the capacitor.
No current flows on the reverse half cycle so the reed switch flies back to discharge the capacitor.
The charge going onto the plates is given by.
I = Q/t
We also know that f = 1/t, so we can combine the two relationships to give:
I = Qf
Þ Q = I/f
Since C = Q/V, we can now write
C = I/fV
A capacitor is connected to a 12volt power supply by a reed switch operating at 400 Hz. The ammeter reads 45 mA. What is the capacitance of the capacitor? 
When we charge up a capacitor, we make a certain amount of charge move through a certain voltage. We are doing a job of work on the charge to build up the electric field in the capacitor. Thus we can get the capacitor to do a job of useful work.
We know that:
1. Energy = charge × voltage
2. Q = CV.
This second relationship tells us that the charge – voltage graph is a straight line.
The capacitor is charged with charge Q to a voltage V. Suppose we discharged the capacitor by a tiny amount of charge, dQ. The resulting tiny energy loss (dW) can be worked out from the first equation:
dW = V × dQ
This is the same as the area of the little rectangle on the graph.
If we discharge the capacitor completely, we can see that:
Energy loss = area of all the little rectangles
= area of triangle below the graph
= ½ QV
By substitution of Q = CV, we can go on to write:
E = ½ CV^{2}
What is the energy held by a 50 000 mF capacitor charged to 12.0 V? 
Many electronic circuits use the charge and discharge of a capacitor. If we discharge a capacitor, we find that the charge decreases by the same fraction for each time interval. If it takes time t for the charge to decay to 50 % of its original level, we find that the charge after another t seconds is 25 % of the original (50 % of 50 %). This time interval is called the halflife of the decay. The decay curve against time is called an exponential decay.
The voltage, current, and charge all decay exponentially during the capacitor discharge.
We can plot a graph using a circuit like this:
We can note the voltage and current at time intervals and plot the data, which gives us the exponential graph shown on the next page. We should note the following about the graph:
Its shape is unaffected by the voltage.
The half life of the decay is independent of the voltage.
The current follows exactly the same pattern as I = V/R.
The charge is represented by the voltage, as Q = CV.
The graph is asymptotic, i.e. in theory the capacitor does not completely discharge. In practice, it does, after about 5 RC.
The graph looks like this:
The graph is described by the relationship:
Q = Q_{0} e^{ –t/RC}
[Q – charge (C); Q_{0} – charge at the start; e – exponential number (2.718…); t – time (s); C – capacitance (F); R – resistance (W).]
For voltage and current, the equation becomes:
· V= V_{0} e^{ –t/RC}
· I = I_{0} e^{ –t/RC}
The product RC (capacitance × resistance) which we see in the formula is called the time constant. It is given the physics code t (t, tau, is a Greek lower case letter‘t’). The units for the time constant are seconds. We can go back to base units to show that ohms × farads are seconds. So if we discharge the capacitor for RC seconds, we can easily find out the fraction of charge left:
V= V_{0} e^{ –RC/RC} = V_{0} e^{ –1} = 0.37 V_{0}
So after RC seconds the voltage is 37 % of the original. This is used widely by electronic engineers. To increase the time taken for a discharge we can:
Increase the resistance.
Increase the capacitance.
We can link the halflife to the capacitance. At the half life:
· Q = Q_{0}/2
· t = t_{1/2}
Therefore:
Q_{0} ÷ 2 = Q_{0} e ^{– t}_{1/2}^{/RC}
Þ ½ = e ^{– t}_{1/2}^{/RC}
Þ 2^{1} = e ^{– t}_{1/2}^{/RC}
Þ e ^{+ t}_{1/2}^{/RC} = 2
Þ log_{e} (2) = t_{1/2}/RC
Þ t_{1/2}_{ } = log_{e} (2) × RC = 0.693 × RC
The halflife is 69 % of the time constant.
Example A 5000 mF capacitor is charged to 12.0 V and discharged through a 2000 W resistor.
(a) What is the time constant? (b) What is the voltage after 13 s? (c) What is the halflife of the decay? (d) How long would it take the capacitor to discharge to 2.0 V? 
Answer (a) Time constant = RC = 2000 W × 5000 × 10^{6} F = 10 s.
(b) Use V = V_{0} e ^{ –t/RC} Þ V = 12.0 V × e ^{– 13 s/10 s} = 12.0 × e ^{– 1.3} = 12.0 × 0.273 Þ V = 3.3 volts
(c) t_{1/2}_{ } = 0.693 × RC = 0.693 × 10 = 6.93 s.
(d) We need to rearrange the formula by taking natural logarithms.
V = V_{0} e ^{–t/RC}
Þ V / V_{0 }= e ^{–t/RC}
Þlog_{e} V  log_{e} Vo = t/RC [When you divide two numbers, you subtract their logs]
Þ 0.693 – 2.485 =  t/10
Þ t/10 = 1.792
Þ +t/10 = +1.792 Þ t = 1.792 × 10 = 17.9 s

A 470 mF capacitor, initially charged up to 12.0 V is discharged through a 100 kW resistor. (a) What is the time constant? (b) What is the voltage after 10 s? (c) How long does it take for the voltage to drop to 2.0 V? 
When an uncharged capacitor is first connected to a supply with a voltage, the electrons can easily move onto the plates.
A good analogy to think of is a commuter train. Before the train starts off on its journey, it is not crowded and it's easy to get on and find a seat. As it goes along the line towards the big city, more and more commuters get on, so it's harder to find a seat. Eventually there are no seats and it's standing room only. And then it gets ever harder even to get on the train...
The greater the charge, the harder it is for electrons to crowd on. We know that:
Q = CV
Therefore the greater the charge, the greater the voltage.
The relationship that describes the exponential increase in voltage is:
Notice the second graph. This tells us that the current goes down exponentially. This is because the current falls as the voltage increases. The relationship for the current is:
We can show that after 1 time constant, the voltage is about 63 % of maximum:
Example A capacitor of capacitor 56 mF is connected to a 12 V supply in series with a resistor of resistance 81 kW. (a) Calculate the time constant. (b) Calculate the current immediately the current is turned on. (c) Calculate the voltage after 10 seconds. (d) Calculate the current after 10 seconds. 
Answer (a) t = 56 × 10^{6} F × 81 × 10^{3} W = 4.536 s [4.5 s to 2 s.f.] (b) I = V/R = 12 V ÷ 81 × 10^{3} W = 0.148 × 10^{3} A ( = 0.148 mA) [0.15 mA to 2 s.f.] (c) V = 12 V × (1  e^{10/4.536}) = 12 V × (1  e^{2.205}) = 12 V × (1  0.110) = 10.7 V [11 V to 2 s.f.] (d) I = 0.148 × 10^{3} A × e^{2.205} = 1.63 × 10^{ A }= 16.3 mA [16 mA to 2 s.f.] 
A capacitor of 470
mF is charged up through a 150 k resistor, with a final voltage of 12 V? 
Remember:
The curved graph is exponential. It intercepts the vertical axis at 100 % of the charge. It is asymptotic, which means that it never hits the horizontal axis (in theory).
The logarithms are natural logarithms, i.e. to the base e (e = 2.718…).
Half life is the time taken for the voltage to fall to 50 %.
Charge, current, and voltage all fall exponentially.
1. Note that, in the example above, the answers in square brackets [] are to two significant figures as the data are to two significant figures. However the data used in the calculations are to more significant figures. Do the rounding at the end.
2. Other curved graphs are NOT exponential.
3. Make sure you use the ln keys on your calculator, not the lg keys.
4. The voltage falls to 25 % after two halflives, NOT zero!