Electricity Tutorial 1  Basic Electrical Measurement
The instruments that are of most use to the physicist and the electrical engineer are the voltmeter and ammeter. With these, we can directly measure:
Potential Difference (or voltage)
Current.
We can then use the data to get:
Resistance (volts ÷ amps);
Power (volts × amps).
The multimeter is often used, as it can measure voltage and current (but not at the same time).
Notice that there is an analogue meter (a meter with a scale) and a digital meter (a meter where the readout is a number). It is important that you learn how to use these instruments correctly in your practical work. If in doubt, ask your teacher.
Current is measured with an ammeter, which is wired in series with the component. The voltmeter is wired in parallel with the component.
The base electrical quantity is current, the flow of charge. All other electrical quantities are derived from it. Current is measured in ampères, or amps (A).
Charge is measured in coulombs
(C), which is defined as:
1
coulomb is the quantity of charge carried past a given point if a steady current
of 1 amp flows for 1 second.
A single
electron carries a charge of 1.6 × 10^{19} C.
1 coulomb is equivalent to 6.2 ´10^{18} electrons. It is much more convenient to use this rather than counting individual electrons. You would buy a 1 kg bag of sugar rather than counting all the crystals in it.
What do you think an electron is? 
Charge
and current are linked by a simple formula:
Charge (C) = current (A)
´ time (s)
In the syllabus the formula is written in physics code as:
The symbol D is Delta, a Greek capital letter 'D', meaning “change in”.
There are some important multipliers for current:
1
microamp (1 mA)
= 1
´ 10^{6}
A
1
milliamp (mA) = 1
´
10^{3} A
These are useful when we are dealing with small currents.
However we must remember to convert to
SI
units for doing calculations.
Watch out for this bear trap!
Show that 1 coulomb is 6.2 ´10^{18} electrons. 

A charge of 1.24 C flows in a period of 0.63 s. What is the current? 
Potential
Difference
Potential
Difference is defined as energy per unit charge.
The unit of potential difference is the volt (V). Using the definition, we can define the volt as Joules per Coulomb.
1 V = 1 JC^{1}.
Potential
difference (V) = energy converted (J)
Charge (C)
In physics code we write:
Potential difference is often referred to as voltage.
Conventional current goes from positive to negative. Electrons carry energy around the circuit; they go from negative to positive. In the early days, physicists didn't know about the electron, which is why they got it all wrong. Correction would require a complex rewrite of the Laws of Physics, a task which noone is likely to be bothered to tackle. So all conventional currents are from positive to negative. All currents are treated as conventional.
Sources of Voltage
The two cells in the parallel combination will last twice as long, since the current will be half that taken from the two series cells.
Batteries are often rated in amphours. A 1 amp hour battery can give out a current of 1 amp for 1 hour. The charge contained is:
Q = 1 A × 3600 s = 3600 C
A cordless drill operates using a 14.4 V battery pack. The battery is rated at 2 amp hours which means that it can deliver a current of two amps for a period of 1 hour. How much energy is held by the battery? 
Resistance in a wire is the opposition of a wire to the flow of electricity. It is caused by collisions between the electrons and the atoms in the wire. The hotter the wire, the more chance there is of a collision. Therefore hot wires have more resistance. The formula for resistance is:
Resistance (ohms) = potential difference (volts)
current (amps)
In physics code we write this as R = V/I
Or more commonly:
V = IR
The unit for resistance is ohm (W). (The curious symbol ‘W’ is Omega, a Greek capital letter long Ō.)
What do you understand by the term resistance? 

Use the circuit below to answer the questions:

Watch out for these bear traps in electrical calculations: 
Measuring Electrical Quantities
We can measure voltage and current using a circuit like this. You will be
familiar with this from GCSE, and you will be expected to set it up as a matter
of course.
The voltmeter is wired in parallel with the component, and the ammeter is wired in series with the components.
Normally we treat the instruments as perfect. This means that:
An ammeter has zero resistance.
A voltmeter has infinite resistance.
In reality an ammeter has a very low, but measurable resistance. Normally we ignore this.
Analogue voltmeters have a high resistance. Normally this has little effect, but if we are measuring high value resistances, the current taken by the voltmeter can alter the result.
A digital voltmeter has a very high resistance indeed, and can be regarded as perfect.
Meters of known resistance (Alevel)
This kind of problem is likely to appear in Alevel papers rather than ASlevel.
An analogue meter can be made to measure voltage or current by adding an external resistor:
An ammeter uses a shunt, which is a low value resistor connected in parallel;
A voltmeter uses a multiplier, which is a high value resistor connected in series.
You can't have the two on at the same time! A typical meter found in a school laboratory is shown below:
Notice that there are separate shunts for AC and DC. There is also a separate multiplier for AC voltage.
The picture shows the shunt resistors and the multiplier resistors in an analogue multimeter:
The same applies to digital multimeters, but the shunts and multipliers are not so easy to identify.
The real voltmeter has a circuit diagram like this:
A perfect voltmeter has an infinite resistance. A digital voltmeter has a very high resistance, and can be considered to be almost perfect. An analogue voltmeter will have a resistance of about 50 kilohms. We model the voltmeter of known resistance as a perfect voltmeter in parallel with a known resistance:
The circuit diagram shows how the meter is MODELLED.
The multiplier is actually wired in series. 
Worked example A voltmeter of resistance 35 kW is used to measure the voltage across a 350 kW resistor. The source providing the current can be modelled as a voltage source of 2.0 V with an internal resistance of 45 kW. (a) What is the voltage across the 350 kW resistor when the voltmeter is NOT connected? (b) What is the voltage that is actually read from the meter? 
Answer We model the circuit like this:
(a) We need to find the total resistance of the circuit: R = 45 000 W + 350 000 W = 395 000 W
Work out the current: I = 2.0 V ÷ 395 000 W = 5.06 × 10^{6} A
Now work out the voltage across the 350 kW resistor: V = 5.06 × 10^{6} A × 395 000 W = 1.77 V = 1.8 V (2 s.f.)
(b) A voltmeter with a known resistance is treated as a perfect voltmeter in parallel with a known resistor. With the voltmeter added our circuit becomes:
Now we need to work out the parallel combination of resistors: R^{1} = (350 000 W)^{1} + (35 000 W)^{1} = 31.4 × 10^{6} W^{1} R = 31818 W
Now our circuit with the perfect voltmeter becomes:
Now we work out the total resistance: R = 45 000 W + 31818 W = 76818 W
Then we can work out the current: I = 2.0 V ÷ 76818 W = 26.04 × 10^{6} A
Finally we work out the voltage read by the meter: V = 26.04 × 10^{6} A × 31818 W = 0.828 V = 0.83 V (2 s.f.)
Note that the rounding appropriate significant figures is done at the last step. 
A perfect ammeter has zero resistance. While we can get a very nearly perfect voltmeter with a digital instrument, the same does not apply to an ammeter. There is always a small resistance to the ammeter. A real ammeter is a meter in parallel with a low value resistor called a shunt. You will sometimes find in engineering textbooks references to motors being shuntwound. This simply means that the rotor coils and the field coils are wired in parallel.
The meter is reading the voltage across the shunt resistance.
We model the ammeter as a perfect ammeter (i.e. one with zero resistance) in series with a resistor of known resistance, like this:
(Challenge  Revise internal resistances of cells before trying this one out) The circuit shows a battery of EMF 20.0 V that has an internal resistance of 0.45 W. It is connected to a 350 W resistor. The potential difference is measured using a voltmeter of resistance 35 kW, while the current is measured using an ammeter of resistance 0.50 W. The circuit is shown in the diagram below:
(a) Show that the current from the cell is about 58 mA. (b) Calculate the reading of the voltmeter, which is known to be accurately calibrated. 
Using these values, there is not a great deal of difference compared to using ideal meters. However with very high value resistors, as shown in the worked example, the differences become more obvious.