Current and Charge

Current

Metal, due to it's structure, has a cubic lattice of atoms and a pool of free electrons surrounding them. In metal, these electrons (charge = -1) move to produce the flow of electric current. In an ionic solution, the ions move to form the current when electricity is passed through the solution. This is called electrolysis. In semiconductors, both electrons and holes move to create the current. N-semiconductors have more electrons, p-semiconductors have more holes. When two differently doped pieces of semiconductor are put together (NP junction) this forms a diode. Then three are put together (NPN or PNP) a transistor is formed.

A current is the rate of flow of charge. Charge is measured in coulombs(C) and has the symbol Q. The equation connecting the three quantities is:

Current = charge / time (I=Qt^-1).
Q = coulombs. I = Amperes t = seconds

Doing this can be confusing when different orders of magnitude are used. e.g.

1. 50mC flows in 10s. What is the current?

I==5mA

2. 50uA flows for 1 minute. What charge has flowed in this time?

Energy Transferred Electrically

Power is measured in Watts (Joules / second). In electricity, power = current x potential difference.
P=IV Watts = Amps x Volts

Example 1

For a 100 watt lightbulb plugged into a 240v mains, work out the current.
P=IV   100 = 240I
I = 100/240 = 0.42A

Example 2

For a 2400 watt kettle plugged into a 240v mains, work out the current.
P=IV 2400 = 240I I = 2400/240 = 10A

Another useful equation is power = work done(energy) / time.

P=W/t
Watts = Joules seconds^-1

Combining the two equations:

work done = pd. x current x time - W=VIt

Joules = volts x amps x seconds

Example 1

In 1 hour, how much energy will the lightbulb and kettle mentioned previously take up?

1. Lightbulb
W = Vit = 240 x 4.2 x (60x60) = 360,000J = 360kJ = 3.6 x 105 J

2. Kettle
W = Vit = 240 x 10 x (602) = 8640000J = 8.64MJ = 8.64 x 106 J

Nature of an electric current.

In a wire, the current is carried by free electrons in the wire.
l=length of wire (m)
A=cross sectional area (m²)
e=charge on an electron = -1.6x10^-19C
n=electron density = number of free electrons / m³
I=current (A)
v=drift velocity=speed of charge carriers (ms^-1)

Deriving the equations:
Derive a value for the current:
Volume of the wire = Al
Number of free electrons in conductor = nAl
Total charge that is free to move = nAle
Time required for all electrons to emerge out of the end of the conductor = l/v

Other variations on the theme:
(drift velocity derived)
(current density (J) = I/A)

Example 1

n = 5×10²⁸ m^-3
e = 1.6×10^-19C
I = 1A
A=1mm² = 1x10^-6m²

v = 1.25x10^-4 ms^-1
v=0.125mms^-1 = 1/8mms^-1
This is a small value - electrons move very slowly.

Potential Difference (pd) and Electromotive Force (emf)

Consider the diagram shown below:

r is the internal resistance of the cell. The pd across the terminals of a cell is called the terminal pd - this is the actual voltage that the cell can put out. In this case, 0.6v is lost in pushing the current through the cell. This is called the internal resistance of that cell.

Consider the circuit again.

E=I(R+r)
E=IR+Ir - This is the equation for EMF

Definition of E and V

E = wd/Q
The EMF is the work done when unit charge flows through it. This work is the energy change of chemical energy into electrical energy.

V = wd/Q
The pd is the work done per unit charge as electrical energy is converted into other forms of energy, especially heat energy.

Kirchhoff’s Laws

1st Law

Definition : The algebraic sum of the currents at any one point is zero.
Explanation : The current going into any one point is equal to the current going out.
Formula : SI=0 (When using this formula, current in is +ve and current out -ve)

3 + 4 - 2 - x = 0 Þ x = 5A

This is really a law of conservation of charge - no charge can be gained or lost at any point in a circuit.

2nd Law

Definition : The algebraic sum of the emfs is equal to the algebraic sum of the pds around any closed loop.
Explanation : The sum of the emfs provided by any cells or batteries in a loop of circuit is the same as the sum of the voltages across the components in the circuit.
Formula : SE = Spd.

Example 1

Consider the following very simple circuit:

E = IR + Ir

r is the internal resistance of the power source, which is usually notated as such to distinguish it from external resistances, which use a capital R.

Example 2

Now consider a more complicated example:

For ACDF :
For BCDE :
For ABEF :
From the first law :

Characteristics

The V-I characteristic of a component shows the current that flows through a particular component for a particular potential difference accross it. It is drawn as a graph, with I on the y-axis and V on the x-axis.

1. Fixed Resistor (at constant temperature)

2. Metal wire at constant temperature.

3. Lamp filament.

4. Semiconductor diode.

The first thing to notice about a diode is that the characteristic is different, depending on which way the pd is applied to it. This means that the x-axis needs to extend in both directions, to show a negative pd as well. The second thing to note is the irregularity of the graph. In the positive region (when the diode is forward biased), there is a very low current for the first few 100mv (the resistance is almost infinite), and then the current leaps to a very high voltage very rapidly. The point at which this change takes place is called the barrier voltage. On the other side of the graph (when the diode is reverse biased), there is only a very small current flow, and again a virtually infinite resistance (typically a few m A) This is called the leakage current of the diode.

5. Thermistor (negative temperature coefficient)

This conducts better with increasing voltage, as this causes the temperature to increase, creating more thermally generated pairs.

Determining the Internal Resistance of a Cell.

The following circuit is used to determine the internal resistance of the cell.

Whilst the resistance is varied, voltage and current readings are taken. The following readings were taken:

Pd (Volts)	Current (mA)
1.49	120
1.50	140
1.49	160
1.49	180
1.48	220
1.47	260
1.46	330
1.45	430
1.42	630

These reading were then plotted onto a graph, with the independent variable (current) on the x axis and the dependant variable (voltage) on the y axis.

From Kirchoff's second law, the following equation can be deduced relating to this circuit:
E = IR + Ir
In this, IR is the Pd that was measured on the voltmeter, and therefore the equation can be rearranged:

E = V + Ir
V = E - Ir
V = E - Ir

As the graph plotted is a straight line graph, it has the equation y = mx + c. This can be substituted into the vales on the graph for this equation:

V = -Ir + e
Y = mx + c

Therefore, y is the voltage (as it is on the graph), x is the current (as on the graph), and the intercept on the y axis, c, is the emf of the cell. Therefore, the internal resistance of the cell is -m, the gradient of the line on the graph. Therefore:

Intercept = c = 1.5V = Emf of the cell.

m = gradient =

The gradient is in ohms because V = IR (Ohm's law), and therefore R = V/I.

The internal resistance = -m = 1.25W.

Examples of Internal Resistance

The small 1.5V cell that is widely used in many applications has an internal resistance of typically 1.25W. This means that the maximum current that the cell can deliver is . This prevents the cell delivering currents so large that they would cause the cell top catch fire, or be flattened instantly.

Car Battery

A car battery is a series of 6 2V accumulators. These are made out of plates of lead and lead sulphate, with an electrolyte of concentrated sulphuric acid. The battery typically has a total internal resistance of 0.01W. This very low internal resistance allows, at just 12V, a current of up to 100A to be delivered to the starter motor of the car to get it started.

EHT supplies

EHT stands for Extra High Tension. These supplies are very high voltage, and supply voltages in the region of 5kV. These very high voltages are used for some experimental purposes. However, these supplies have a very high internal resistance, typically 5k. This limits the maximum current that they can produce to about 1A, making them much safer to use, and preventing damage to the internals of the system by overheating.

Resistivity

Resistivity is a measure of how much a material opposes the current flowing through it. It is defined as the resistance of unit area of the material per unit length. This leads to the following equation:

, where the units are ohm meters.

To measure this, the resistance of a length of wire is measured. The length of the wire is measured, and the area of the cross section is measured with a micrometer. This should be put across the wire in two directions, and at two different places on the length of the wire, to check that it is homogenous. The results are then substituted into the formula above.

Typical values for r are 10^-8 for good conductors, and 10¹² for insulators. Semiconductors would fall somewhere between these two values.

Resistors in Series and Parallel

Derivation 1 - Resistors in Series

1. In a series circuit the current, I, is the same at all points in the circuit.
2. In a series circuit, V_T = V₁ + V₂ + V₃
3. From Ohm's law, V = IR

Derivation 2 - Resistors in Parallel

From this, given sufficient information, any values in circuits with resistors in series and parallel can be worked out.

Heating effect of a current (Ohmic heating)

Metal Wires.

The resistance causes the current flow to drop. As e and n are constants, and A increases with temperature, increasing the current flow, it follows that v must go down. This means that the speed of the electrons goes down as the temperature goes up, because the anions vibrate, inhibiting their progress.

Insulators

There are no free electrons at all in an insulator (e.g. glass) and therefore an electric current can not flow through an insulator.

Semiconductors

At any temperature above absolute zero thermally generated pairs are produced, leading to there always being some free electrons in a semiconductor. As the temperature is increased, more pairs are produced. This in turn reduces the resistance and therefore increases the current, by increasing n in the equation. This can lead to the possibility of thermal runaway, which is where the current flowing through the semiconductor and the temperature of the semiconductor both increase, leading to damage to the device.

Ionic solutions

In an ionic solution, ions move instead of electrons. They can have a &plusmin;2 or &plusmin;3 charge, leading to an increase in q, the charge on the charge carrier (the equation is really I = nAvq)

Potential Dividers

What is the output voltage, Vout, of the following circuit?

• V_out = IR₂
• V_in = I(R₁+R₂)
• 
• 

Advantages of a potential Divider

It allows us to have an output voltage of anywhere between 0 and Vin.
The limits of the system are as follows, and are defined by the nature of the expression :
If R₁ is very small, V_out = V_in
If R₁ is very large, V_out = 0
If R₁ = R₂, V_out = 1/2 V_in

This type of circuit is useful for plotting the characteristic of a diode, where precise measurements are needed.

Using sensors with the Potential Divider:

1. Using a thermistor:

When T is high, V_out=V_in.
When T is low, V_out will be small.

2. Using an LDR

With an LDR, Light Decreases Resistance

In the dark, V_out = 0

In the light, V_out = V_in.

This type of circuit would be used in an umpire's light meter, where the output needs to be proportional to the light level.

If the LDR and the resistor were swapped round, then V_out would be high in the dark. This would be the type of circuit used in a sensor on a streetlight.

Capacitors

Capacitors store charge, therefore they store energy, since V=w/Q. Capacitance is defined as the charge stored per volt of pd applied across the plates, therefore

The following circuit is one used to charge up a capacitor:

When the switch is closed, the -ve terminal of the cell repels electrons which move onto the right-hand plate of the capacitor. This repels an equal amount of charge on the opposite plate, which moves towards the +ve terminal of the cell. This process stops when the voltage across the capacitor is the same as the voltage that is being applied to it (and so the capacitor is pushing more electrons away with the same force as they are being pushed on).The charge that is stored is actually stored on both places, but only flows round once. In the discharge circuit, the electrons are repelled off the +ve plate, and flow round the circuit before neutralising the +ve charge on the other plate.

To measure the capacitance of the capacitor, the following circuit is set up:

R is varied so as to keep the ammeter reading constant. This allows us to calculate the charge flowing onto the plates, and therefore allows Q to be calculated. A graph is plotted of the voltage against the charge, and the gradient, dQ/dV, is equal to the capacitance of the capacitor.

Capacitor Calculations.

Capacitors in Series

1. In the circuit shown above, Q, the charge on each capacitor in the circuit, is the same.
2. In a series circuit, voltages add. Therefore, V_T = V₁ + V₂ + V₃.
3. Q, the charge on each capacitor will be the same, even if the capacitances are different. Therefore, V₁, V₂, and V₃ will be different
4. Taking the equation
   
   and substituting it into the above equation,
   .
5. Cancelling by Q,
   .

Capacitors in parallel

1. In this circuit, the voltage across each capacitor is equal, and equal to the supply voltage.
2. In the capacitances of the capacitors are different, the charge on the capacitors will be different.
3. The total charge stored, therefore, = Q₁ + Q₂ + Q₃.
4. Substituting Q = CV, .
5. Cancelling by V, .

When resolving problems with capacitors, look for sets of capacitors for which an equivalent value can be calculated. Simplify until there is only one equivalent capacitor left.

The Energy Stored in a Capacitor

Consider a capacitor being charged at constant current.

Imagine an increase in charge from Q to Q+DQ, where DQ is very small. This happens whilst the voltage is at V.

Therefore, the area of the strip = VDQ. The units of this will be joules, because wd = VQ.

Therefore, the area under the graph will be the total work done, i.e. the energy stored. The area of the graph = ˝VQ. Q=CV, and therefore wd = ˝VCV = ˝CV˛.