Presentation on theme: "Series and Parallel Circuits. In series circuits, the current passes through each component in turn. In parallel circuits the current splits, some."— Presentation transcript:
Series and Parallel Circuits
In 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
V1V1 V2V2 V3V3 A R1R1 R2R2 R3R3 I 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:V tot = V 1 + V 2 + V 3 From Ohm’s Law we know: V tot = IR tot ; V 1 = IR 1 ; V 2 = IR 2 ; V 3 = IR 3 IR tot = IR 1 + IR 2 + IR 3 R tot = R 1 + R 2 + R 3 This is true for any number of resistors in series.
A1A1 A2A2 A3A3 R1R1 R2R2 R3R3 I tot A t ot I1I1 I2I2 I3I3 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.
Therefore I tot = I 1 + I 2 + I 3 From Ohm’s Law, I = V/R, we can write: I tot = V ; I 1 = V; I 2 = V; I 3 = V R tot R 1 R 2 R 3 V = V + V + V R tot R 1 R 2 R 3 1 = R tot R 1 R 2 R 3 This is true for any number of parallel resistors.
Kirchhoff I I 1 + I I 3 = 0 I = 0 The symbol 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 II Potential differences add up to the battery voltage. Electricity does not leak from wires!
V1V1 V2V2 A R1R1 R2R2 I E A B C I1I1 I2I2 I3I3 Kirchhoff I Kirchhoff II
Let 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 IR 1 volts From B to C the p.d. drop is IR 2 volts From C to A the pd. change is – E volts. If we add up all the voltages, we can write: IR 1 + IR 2 = E This is another way of saying that the voltages add up to the battery voltage.
In its simplest form it is two resistors in series with an input voltage V s across the ends. An output voltage V out is obtained from a junction between the two resistors. R1R1 R2R2 V out VsVs 0 V
If the output current is zero, the current flowing through R 1 also flows through R 2, because the resistors are in series. So we can use Ohm’s Law to say: I = V s R 1 + R 2 Now V out = IR 2 = V s __ × R 2 R 1 + R 2 V out = R 2 ___× V s R 1 + R 2 This result can be thought of as the output voltage being the same fraction of the input voltage as R 2 is the fraction of the total resistance. There is no need to work out the current.
Potential dividers are used in inputs to electronic circuits The sound of silence