Series and Parallel Circuits

Transcription

1 Series and Parallel Circuits Components in a circuit can be connected in series or parallel. A series arrangement of components is where they are inline with each other, i.e. connected end-to-end. A parallel arrangement of components is where they are connected across each other where the current has more than one path through that part of the circuit. Measuring Current and Potential Difference or oltage Electric current is measured using an ammeter which is connected in series with the component. Potential difference (p.d.), or voltage, is measured using a voltmeter which is connected in parallel with the component. Current and oltage (p.d.) Series Circuits The current is the same at all points in a series circuit. The sum of the potential differences across the components in a series circuit is equal to the voltage of the supply.

2 Current and voltage (p.d) Parallel Circuits The potential difference across components in parallel is the same for all components. The sum of the currents in parallel branches is equal to the current drawn from the supply. Resistors in Series When more than one component is connected in series, the total resistance of all the components is equivalent to one single resistor, R T, calculated using the relationship Resistors in Parallel When more than one component is connected in parallel, the total resistance of all the components is equivalent to one single resistor, R T, calculated using the relationship

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4 Example mixed series and parallel circuit Calculate the total resistance of this circuit ) Work out which resistors are in series and which are in parallel Simplify one step at a time 2) Deal with series combination R 2 and R 3 first R T = R 2 + R 3 = = 2 Ω 3) Simplified version looks like this The combined resistance, R A of 2 Ω is in parallel with resistor R 4 This combination is in series with resistor R = which means 4) This is the same as R T = R + R comb = = 2 Ω A single resistance of 2 Ω is equivalent to the 4 resistor combination.

5 Advantages of parallel circuits If one component stops working the rest keep working Can have combinations of components switched on and off as desired All components connected in parallel will have the same voltage across them o Connecting components to the mains supply of 230 means that all components receive the full 230 Potential divider circuits A potential divider is a device or a circuit that uses two (or more) resistors in series, or a variable resistor (potentiometer) to o provide a fraction of the available voltage (p.d.) from the supply. The supply voltage is shared (divided up) between the components The larger resistance takes the larger share of the supply voltage in direct proportion o The ratio of resistances = the ratio of voltage shared between the resistors

6 Example 2 Consider the following voltage divider circuit: 20 kω 2 Y 0 kω X (a) Calculate the p.d. across X Y. R = 0 kω R 2 = 20 kω R R R 2 S S = 2 =? (b) Another 0 k resistor is added as shown. Calculate = 4 the new p.d. across X Y. 20 kω 2 0 kω Y 0 kω X R = 5 kω R 2 = 20 kω S = 2 =? R R R 2 S = 2.4

7 Electrical Energy and Power When there is current in a component, an energy change takes place. o e.g. when there is current in the filament in a lamp, the electrical energy is converted to heat and light. o In an electrical heating element, the energy change takes place in the resistance wire Electrical energy, like all types of energy is measured in Joules. Power is the rate of energy transfer (same definition as on page 4 for mechanical power) Energy (J) Power (W) time (s) The units of power are Watts (W). Watt means Joule of energy is transferred every second. Power, current and voltage The power (P) is also equal to the current (I) in Amps multiplied by the voltage () in volts. voltage () power (W) current (A) Example An electric heater operates on the mains voltage of 230. The heating elements operate at a current of 8 A. Calculate the power developed by the heater. = 230 I = 8 A P =? P = I = 8 x 230 = 840 W

8 More Power equations The equations P = I and = IR can be combined by substitution to form two other equations. P = I 2 R P = I and substitute = IR into the equation for gives P = I IR = I 2 R P = 2 /R P = I and substitute I = into the equation for I gives P = x = This means there are three expressions for the power which are equivalent to each other Example An electric heater operates on the mains voltage of 230. The heater has a power rating of 2 0 kw. Calculate the resistance of the heating element of the heater. P = 2 kw P = R = = 230 R =?

9 Example 2 In the circuit shown opposite calculate: (a) the current in each resistor 40 (b) the p.d. across each resistor (c) the power dissipated in each resistor. 4 Ω 6 Ω 5 Ω 0 Ω (a) R S = R + R 2 = = 0 Ω R p R P = 5 Ω R T = = 0 Ω I 40 T T 4 A (current through 5 Ω) RT 0 The resistance of each parallel section is 0 so current through 6, 4 and 0 resistors all = 2 A (b) For the 5 Ω resistor: = IR = 4 5 = 20 for the 4 Ω resistor: = IR = 2 4 = 8 for the 6 Ω resistor: = IR = 2 6 = 2 for the 0 Ω resistor: = IR = 2 0 = 20 (c) For 5 Ω: P = I 2 R = = 80 W for 4 Ω: P = I 2 R = = 6 W for 6 Ω: P = I 2 R = = 24 W for 0 Ω: P = I 2 R = = 40 W