2Key IdeasIn series circuits, the current passes through each component in turn.In parallel circuits the current splits, some flowing through one branch, and some through another.We can find a single resistor equivalent to these circuits.Kirchhoff’s Laws can be used to analyse circuits
3Series Circuit We know two things about this circuit: V1V2V3AR1R2R3IWe 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.
4Therefore:Vtot = V1 + V2 + V3 From Ohm’s Law we know: Vtot = IRtot; V1 = IR1; V2 = IR2; V3 = IR3 IRtot = IR1 + IR2 + IR3Rtot = R1 + R2 + R3 This is true for any number of resistors in series.
5Parallel Circuit For a parallel circuit we know two things: ItotAtotI1I2I3For 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.
6From Ohm’s Law, I = V/R, we can write: Therefore Itot = I1 + I2 + I3From Ohm’s Law, I = V/R, we can write:I tot = V ; I1 = V; I2 = V; I3 = VRtot R R R3 V = V + V + VRtot R R R3 1 =Rtot R1 R2 R3This is true for any number of parallel resistors.
7Kirchhoff’s Laws Kirchhoff II Kirchhoff I I1 + I2 + - I3 = 0SI = 0The symbol S is Sigma, a Greek capital letter ‘S’, which means ‘sum of’.The total current flowing into a point is equal to the current flowing out of that point.Kirchhoff IIPotential differences add up to the battery voltage.Electricity does not leak from wires!
8Using Kirchhoff I & II Kirchhoff I Kirchhoff II I1 I3 I2 E C A I R1 R2 V1V2AR1R2IEBC
9Let us do a journey around the circuit from A to B to C, and back to A. · From A to B the p.d. drop is IR1 volts· From B to C the p.d. drop is IR2 volts· From C to A the pd. change is – E volts.If we add up all the voltages, we can write:IR1 + IR2 = EThis is another way of saying that the voltages add up to the battery voltage.
10Potential DividerIn its simplest form it is two resistors in series with an input voltage Vs across the ends.An output voltage Vout is obtained from a junction between the two resistors.R1R2VoutVs0 V
11 I = Vs R1 + R2 Now Vout = IR2 = Vs__ × R2 Vout = R2___× Vs If the output current is zero, the current flowing through R1 also flows through R2, because the resistors are in series. So we can use Ohm’s Law to say:I = VsR1 + R2Now Vout = IR2 = Vs__ × R2 Vout = R2___× VsThis result can be thought of as the output voltage being the same fraction of the input voltage as R2 is the fraction of the total resistance. There is no need to work out the current.
12Use of the Potential Divider The sound of silencePotential dividers are used in inputs to electronic circuits