Use these quick notes to help you revise each topic from the Chapter. You can also download a PDF with revision cards to print double sided and 2 slides per sheet to get small cards to play revision gamesAnimated Science Physics AS Revision Cards Chapter 5 but the information is the same as on this post!
5.1 Circuit Rules
Kirchhoff’s Current Rule
The “current law” states that at a junction all the currents should add up.
I3 = I1 + I2 or I1 + I2 – I3 = 0
- Current towards a point is designated as positive.
- Current away from a point is negative.
- In other words the sum of all currents entering a junction must equal the sum of those leaving it.
- Imagine it like water in a system of canals!
There are some important multipliers for current or other electrical quantities:
1 microamp (1 μA) = 1 x 10-6 A
1 milliamp (mA) = 1 x 10-3 A
Also remember to make sure you work out current in Amps and time in seconds in your final answers!
Kirchhoff’s Voltage Law
For any complete loop of a circuit, the sum of the e.m.f.s equal the sum of potential drops round the loop. A cell is a positive e.m.f. a resistance is a negative.
ε = V1+V2 +….
PD Series or Parallel
In a series circuit the energy is shared between components according to the Kirchoff law.
In a parallel circuit each branch is at the PD of the power supply. Then on that branch the PD is shared as with a series circuit.
Current Series or Parallel
Current is a series circuit is the same everywhere.
Current in a parallel circuit splits at branches according to resistance and recombines at a later junction.
Ammeters count the flow of electrons or Cs-1 in a circuit. They have a very low resistance and only take a very tiny current to make them work. This is why we place them in series with components.
A voltmeter (can be an oscilloscope as well) is a very high resistance meter and simply compares one side of a component to another. It tells us the energy potential difference or PD. This is measured in volts or JC-1. This is why they are placed in parallel with components.
You should know the symbols for al the basic items such as… filament lamp, resistors of every type, battery, cell, LDR, Diode, LED, Thermistor, heater, motor, A,V, Internal resistance inside a cell, AC supply etc..
5.2 More on Resistance
When you connect a resistor in series with another the current flows through both. This increases the resistance overall. The total resistance of a branch is easy to find.
RT = R1 + R2 + ….
15 Ω = 10 Ω +5Ω
When you connect a resistor in parallel with another the current flows through both. This decreases the resistance overall as there are many branches. The total resistance of a branch is harder to find.
1/RT = 1/R1 + 1/R2 + ….
3/10 Ω = 1/10 Ω +1/5Ω = 10 Ω /3 = 3.3 Ω
Charge carriers transfer kinetic energy to positive ions through repeated collisions. The pd across the material then provides an accelerating force to the charge carrier which then collides with another positive ion. The heating effect is energy delivered or energy delivered per second…
P=VI, P = I2R = P V2/R
Working out circuits
- Combine the resistances on each branch using parallel or series equations.
- Work out the current in each branch.
- Work out the p.d. across branch.
- Work out the energy dropped across a component.
5.3 Internal Resistance
This is the electromotive force or push produced by an energy source… ε = E/Q. It is measured in volts (JC-1)
PD Across Terminals
When you actually start to draw current from an energy source the emf drops the more you draw current. We lose energy in the power supply. Hence e is only theoretical and we actually get a PD less than this for our real circuit.
If we think of the power source of having an internal resistance to current flow the sum of all the emf in the circuit..
ε = Vint+Vload
ε = Irint+IRload
ε = I(rint+Rload)
ε – Irint = Vload
IR Graphing 1
Vload =ε – Ir
This shows us that when we make a simple circuit and then draw more current through a reduced resistance the internal resistance consumes more of the e when the current is higher.
Hence a graph shows intercept as “ε” and the gradient as “r”.
Int Res Series and Parallel
If you have cells in series their internal resistance adds up to be…
Series -> rT = r1 + r2 + ….
Parallel -> 1/rT = 1/r1 + 1/r2 + ….
If graphing the circuit you would need to treat the gradient as rT.
Power and EMF
We can also think about the situation for internal resistance as a power transfer and then rearrange our equations to this…..
Pcircuit = Pcell + Pload
P= εI = I2r + I2R
If we consider that P = I2R = I2(R+r) then we can plot a complex equation curve to look at how Power varies when you compare the internal resistance to the load. We should know that when r=R i.e. 4Ω = 4 Ω then the maximum power will be delivered to the circuit.
IR Graphing 2
We can think of our formula in the way of a straight line graph….
Vload = (-r)I + ε
y = (m)x + c
Vload = y
-r = gradient
I = x
ε = c
5.4 More Circuit Calculations
If you have a circuit with two sources of emf in series they either add up or subtract depending on direction.
If the cells are in parallel they will give the same PD as one but increased current flow.
If we consider a diode in a circuit it will share PD or other energy with components depending on the overall voltage as the resistance of the diode will change. You can have several different circumstances. Simply treat them as new situations.
Are made from a PN sandwich which means we can separate charged electrons and produce an electric field. If a light photon through a glass screen falls onto our sandwich it releases the electrons to move through the field and produce an electric current. This is how a solar cell works.
Diodes and AC
If you place a diode into a direct current circuit then it will conduct in a forwards direction. However in an AC circuit will turn off the current in one direction and create half waves with missing half waves.
5.5 Potential Dividers
Simply divides the energy in a circuit but the current through the main circuit is constant.
Allows us to provide a variable PD to a component due to a change in condition i.e. pressure, temp, light, position.
In the most basic form of a two resistor potential divider the current is the same throughout the circuit i.e. VS = IR. Then there is a V dropped over R1 and R2. The % drop across each is calculated by this formula which you must memorise.
V2 = R2 /( R2+ R1) x Vs
This changes to R1 on top for V1.
Pot Div Issues
Voltage: Pot div circuit can provide the full range of voltage from V -> 0V, while a variable resistor circuit will not reach 0V.
Current: In a pot div circuit the load resistance (bulb) is in parallel with the variable resistor which means that the overall resistance is less and more current flows in the circuit.
Energy: In a pot divider circuit the current flow is more as there are two pathways for current to flow. Hence energy flow is more!
Ratios for V and R
We find that the energy splits up according the ratio of resistances so for R1 and R2….
V1/V2 = R1/R2
This can be very useful if you don’t have the current in a circuit and need a quick fire answer. But you have to memorise this one!