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Published byDestiny Hill Modified about 1 year ago

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Series and Parallel Circuits

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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

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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.

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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.

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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.

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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 = 1 + 1 + 1 R tot R 1 R 2 R 3 This is true for any number of parallel resistors.

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Kirchhoff I I 1 + I 2 + - 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!

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V1V1 V2V2 A R1R1 R2R2 I E A B C I1I1 I2I2 I3I3 Kirchhoff I Kirchhoff II

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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.

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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

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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.

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Potential dividers are used in inputs to electronic circuits The sound of silence

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