| 2. Solar Cells and Electrical Circuits |
For any electrical circuit to work, it must:
Any source of potential difference must be able to separate charge, so that:
Remember that protons never move. The electrons move from negative to positive. However conventional current flows from positive to negative; the early physicists got it wrong. Unless advised otherwise, treat all currents as conventional.
Charge separation occurs by:
In satellites the power source is usually a solar cell, the symbol for which is:
However we will represent power sources in these notes as normal cells or batteries:
We can combine cells in different ways:
In series, in which the total voltage is the sum of the voltage of all the cells. The current is limited to the maximum current that the cell can produce.
In parallel, in which the current is the sum of the currents delivered by all the cells, but the voltage is the same as each single cell.
A combination of the two.
If we want a higher voltage, we put the cells in series. If we want a big current, we put them in parallel.
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.
1 electron carries a charge of 1.6 × 10-19 C.
1 coulomb is equivalent to 6.2 × 1018 electrons.
Charge and current are linked by a simple formula:
Charge (C) = current (A) × time (s)
In Physics code:DQ = IDt
There are some important multipliers for current:
1 mA = 1 × 10-6 A
1 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.
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:
V = W
Q
Potential difference is often referred to as voltage.
Since potential difference is defined as energy per unit charge, we can rearrange things to state:
Energy transferred (J) = charge transferred (C) × potential difference (V)
In physics code:
DE = DQV
Since
charge (C) = current × time
DQ = IDt
we can substitute DQ and get:
DE = VIDt
In other words:
Energy (J) = voltage (V) × current (A) × time (s)
Resistance is the opposition to the flow of an electric current. Resistance in a conductor is thought to arise due to the collisions between the charge carriers and the ions in the lattice. The internal energy rises, so the conductor gets hot. The hotter the conductor, the greater the probability of a collision between an ion and an electron. The resistance in hot conductors rises.
Resistance (W) = Potential difference (V)
Current (A)
In physics code:
The unit for resistance is ohm (W). (The curious symbol ‘W’ is Omega, a Greek capital letter long Ō.)
Resistance is the ratio of the voltage to the current, described in the simple equation R = V/I. In a metallic conductor, we find that if we alter the voltage or the current, the other variable changes in such a way that the ratio remains constant.
This is Ohm’s Law, which states:
The current in a metallic conductor is directly proportional to the potential difference between its ends provided that the temperature and other physical conditions are the same.
A conductor that obeys Ohm’s Law is called an ohmic conductor.
We can easily measure voltage and current, using the data to plot voltage current graphs. We normally put the voltage on the y-axis and current on the x-axis. This allows us to determine the resistance from the gradient.
This is a voltage current graph for an ohmic conductor:
The straight line shows a constant ratio between voltage and current, for both positive and negative values. Ohm’s Law is obeyed.
For a filament lamp we see:
The resistance rises as the filament gets hotter, which is shown by the gradient getting steeper.
For a diode:
The diode starts to conduct at a voltage of about +0.6 V. Then the current rises rapidly for a small rise in voltage. If the current is reversed, almost no current flows until the breakdown voltage is reached.
In a series circuit, the electrons in the current have to pass through all the components, which are arranged in a line. Consider a typical series circuit in which there are three resistors of value R1, R2, and R3. The values may be the same, or different.
We know two things about this circuit:
All the voltages add up to the voltage given out by the battery.
The current is the same all the way round.
Therefore:
VT = V1 + V2 + V3
From Ohm’s Law we know:
VT = IRT; V1 = IR1; V2 = IR2; V3 = IR3
Þ IRT = IR1 + IR2 + IR3
Þ RT = R1 + R2 + R3
This is true for any number of resistors in series.
Parallel circuits have their components in parallel branches so that an individual electron can go through one of the branches, but not the other two. The current splits into the number of branches there are:
In this case, the current will split into three.
For a parallel circuit we know two things:
The voltage across each branch is the same
The currents in each branch add up to the total current.
IT = I1 + I2 + I3
From Ohm’s Law, I = V/R, we can write:
I T = V ; I1 = V; I2 = V; I3 = V
RT R1 R2 R3
Þ V = V + V + V
RT R1 R2 R3
Þ 1 = 1 + 1 + 1
RT R1 R2 R3
This is true for any number of parallel resistors.
Remember to invert the 1/Rtot to get the resistance in ohms
We can combine resistors in both series and parallel. Tackle the problem step by step.
Work out the total resistance of the parallel combination.
Work out the total resistance of the circuit by adding your answer in the previous step to the values of the series resistors.
Take care with such problems:
Make sure that the voltages across each part of the circuit add up to the battery voltage.
Make sure the currents in the parallel part of the circuit add up to the battery current.
If they don’t, go back and check what you’ve done wrong!
A battery is said to produce Emf or electromotive force which is defined as
the energy converted into electrical energy when unit charge passes through the source.
It represents the total energy that can be supplied to a circuit. No circuit at all is 100 % efficient. Some energy is dissipated in the wires, or even in the battery itself.
The curly 'E' is the physics code for emf, and the units are Joules per coulomb (JC-1) or volts (V).
A more simple and practical way of remembering emf is to say that it is the terminal voltage of a battery or generator in open circuit, i.e. when no current is being drawn.
All batteries and generators dissipate heat internally when giving out a current, due to internal resistance. A perfect battery has no internal resistance, but unfortunately there is no such thing as a perfect battery. Nickel-Cadmium and Lead-Acid batteries have very low internal resistance, and we can regard these as almost perfect. These batteries can provide very high currents.
A perfect voltmeter has infinite resistance. A digital multimeter has a very high resistance, so needs a tiny current; it is almost perfect. An ordinary moving coil voltmeter has a relatively low resistance, so it takes a small but appreciable current.
Suppose we now add a load. We will assume the wires have negligible resistance.
This time we find that the terminal voltage goes down to V. Since V is less than ℇ, this tells us that not all of the voltage is being transferred to the outside circuit; some is lost due to the internal resistance which heats the battery up.
Emf = Useful volts + Lost volts
In physics code:
ℇ = V + v
So we can represent the circuit as:
So our cell is now a perfect battery in series with an internal resistor, r. You cannot open up the battery to find the internal resistor; it is part and parcel of the battery.
We can now treat this as a simple series circuit and we know that the current, I, will be the same throughout the circuit. We also know the voltages in a series circuit add up to the battery voltage.
Emf = voltage across R + voltage across the internal resistance
ℇ = V + v
We also know from Ohm’s Law that V = IR and v = Ir, so we can write:
ℇ = IR + Ir
which becomes
ℇ = I(R + r)
Many students panic at the sight of internal resistance problems. All you have to do is turn the cell with the internal resistance into a perfect battery in series with its internal resistor, and treat it as a simple series circuit.
Suppose that the charge that flowed through an electrical component was in the form of a steady current that flowed for t seconds. The charge was converted to DE joules.
We know that DQ = IDt and DE = DQV.
Now:
Power (W) = energy (J)
time (s)
In Physics code:
P = VIDt
Dt
The
Dt
terms cancel out so we are left with:
Power is measured in watts (W). 1 watt = 1 joule per second
We can relate the power to resistance.
We know that
P = VI and V = IR.
So
it doesn’t take a genius to see that by substituting the second equation into
the first, we get:
P = I2R
By a similar process we can write:
P = V2
R
The maximum power transfer occurs when the load = internal resistance. This is shown on the graph below. In this case the internal resistance is 1 ohm, and the load resistance is 1 ohm.
The power in the load is 195 W. Therefore the power dissipated in the internal resistance is 195 W.
Since:
Efficiency = useful power × 100 %
total power
we can see that the efficiency at maximum power deliver is 50 %.
The load-efficiency graph is:
If we look at where the load is the same as the internal resistance (1 ohm) we see that the efficiency is 50 %. This means that a lot of energy is being wasted in heating up the battery. If a lot of power is being transferred, the battery will get hot and could be damaged.
If the load resistance is higher than the internal resistance, the efficiency goes up. To achieve maximum efficiency, the load resistance needs to be much higher than the internal resistance.
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