Higher Physics Electricity.

Higher Physics Electricity

Electricity: Contents Monitoring and measuring a.c. Current, p.d., power and resistance Electrical sources and internal resistance. Capacitors Conductors, semiconductors and insulators p-n junctions

Electricity: Monitoring and measuring a.c.
Alternating and direct current When a circuit is connected to a battery, the current always flows round the circuit in one direction. This is called direct current (d.c.)

Mains Electricity: Monitoring and measuring a.c.
When a circuit is connected to the mains the current flows round the circuit in one direction and then in the opposite direction. This is called alternating current (a.c.)

An oscilloscope or CRO is used to study electrical signals.

Measuring Peak Voltages Height of peak voltage = Voltage gain setting = Peak voltage =

Measuring Frequency Length of 1 wave = Time base setting = Period = Frequency = 1/Period =

Measuring a.c. signals peak voltage = height x voltage gain (V) 1 1 frequency = (Hz) = period length x time base

Example A signal from an a.c. supply is displayed on an oscilloscope The time-base is set at 0·01 s/div. The Y-gain is set at 4·0 V/div. Calculate the peak voltage and the frequency of the signal. period = length x time-base = 4 x 0·01 = 0·04 s peak voltage = height x voltage gain = 3 x 4 = 12 V frequency = = 25 Hz 1 0·04

Peak and R.M.S. Voltage The voltage from an a.c. supply is always changing. The R.M.S. voltage of an a.c. supply is the voltage that is equivalent (produces the same effect) as a d.c. supply of that value. Experiment - comparing a.c. and d.c. supplies. switch voltage from a.c. supply voltage from d.c. supply to CRO Value of d.c. supply = Peak voltage of a.c. supply =

Vpeak = Vrms Also Ipeak = Irms

Example The r.m.s. voltage from an a.c. supply is 6 V. Calculate the peak voltage. Vpeak = Vrms = x 6 = 8·5 V

Electricity: Current, p.d., power and resistance.
Charge, Current and Time An electrical current is a flow of (negative) charge around a circuit. The amount of charge that flows around a circuit depends on the size of the current and the time the charge flows. current in amperes (A) Q = I t time in seconds (s) charge in coulombs (C)

Resistance in Circuits The current in a component in a circuit can be measured by placing an ammeter in series with the component and the potential difference (p.d.) across a component measured by placing a voltmeter in parallel with it. A V A component is said to have a resistance if a potential difference (p.d.) is needed to drive a current through it.

Experiment – Ohm’s Law Aim : To find the relationship between the current in a resistor and the p.d. across it. p.d. (V) current (A)

p.d. (V) current (A)

Potential Difference, Current and Resistance current ( A ) V = I R p.d. ( V ) resistance ( W )

Example What is the potential difference across a 2·0 W resistor if the current through the resistor is 5·0 A? V = IR = 5 x 2 = 10 V

Power tells us how much energy is used each second. e.g. A 26 watt lamp uses 26 joules of energy each second! energy ( J ) time ( s ) power ( W )

If the current through a component is I and the potential difference across a component is V, then the power is given by P = I V . Experiment Aim: To find the power rating of lamps Lamp rating (W) Voltage (V) Current (A) Current x Voltage

Series Circuits Vs I I V2 V1 I The current is the same at all points The sum of the voltages across all the components is equal to the supply voltage. Vs = V1 + V2

Parallel V V Is I1 I2 V The sum of the currents in parallel branches is equal to the current from the supply. Is = I1 + I2 The voltage is the same across all components in parallel

Potential Divider Circuits R1 Vs R2 Voltage output (V2) The voltage (V1) across resistor (R1) is found from V1 = VS - V2

Example Calculate the reading on the voltmeter. 6 W 5 V 4 = x 5 = 2 V 4 W 10 V

LDR in a Potential Divider Circuit As the light level increases the resistance of the LDR decreases and so the voltage across it will R decrease. Vs Output voltage

Thermistor in a Potential Divider Circuit As the temperature of the thermistor increases its resistance decreases and so the voltage across it will R decrease. Vs Output voltage

Example A circuit is set up as shown. 12 V V 10 W 40 W 20 W 60 W Calculate the reading on the voltmeter. 10 ohm 20 ohm reading = 3 -2·4 = 0·6 V 10 20 = x 12 = 2·4V = x 12 = 3 V 50 80

Electricity: Electrical sources and internal resistance
E.M.F. A battery provides an electromotive force (e.m.f.) to drive current around a circuit. The e.m.f is defined as the electrical energy supplied to each coulomb of charge that passes through the battery. The units of e.m.f are the same as p.d. 1 JC-1 = 1 V The e.m.f. of a battery can be measured using a voltmeter to find the p.d. across the terminals of the battery when no current is flowing (open circuit).

Experiment Aim: To investigate the terminal p.d. across a battery as the current changes. V Current (A) p.d. (V) A

p.d. (V) e.m.f. ‘lost’ volts Current (A) As the current increases the terminal p.d. decreases. The battery has an internal resistance r so when a current I passes through the battery some voltage is dropped across it, ‘lost’ volts = I r

The terminal p.d. will always be less than the e.m.f. when a current flows. E = V + I r internal resistance ( W ) e.m.f. ( V ) terminal p.d. ( V ) current ( A )

Example The e.m.f. of a battery is 12 V. When a 8 W resistor is connected to the battery a current of 1 A flows in the circuit. What is meant by an e.m.f. of 12 V ? Calculate the p.d. across the 8 W resistor Calculate the ‘lost’ volts Calculate the internal resistance. 12 V r 8 W When the current is zero, the p.d. across the battery is 12 V. (c) 12 – 8 = 4 V (b) V = IR = 1 x 8 = 8 V (d) V = Ir 4 = 1 x r r = 4 Ω

Example The e.m.f. of a battery is 9 V. When a 6 W resistor is connected to the battery a current of 1·2 A flows in the circuit. Calculate the p.d. across the 6 W resistor Calculate the ‘lost’ volts Calculate the internal resistance. 9 V r 6 W (a) V =IR = 1·2 x 6 = 7·2 V (b) 9 - 7·2 =1·8 V (c) V = Ir 1·8 =1·2 x r r = 1·5 Ω

Example A battery has an e.m.f. of 12 V and internal resistance 2 W. A 8 W resistor is connected to the battery. Calculate the current in the 8 W resistor. (b) Calculate the p.d. across the 8 W resistor. 12 V 2W 8 W total resistance = 8 + 2 = 10 Ω V = IR 12 = I x 10 I = 1·2 A V = IR = 1·2 x 8 = 9·6 V

Example The following circuit is set up to determine the e.m.f. and internal resistance of a battery. Voltmeter and ammeter readings were taken and a graph drawn of the results.

p.d. / V State the em.f. of the battery Calculate the internal resistance of the battery. 7 6 5 4 3 2 1 current/ A 1 2 3 4 5 (a) 6 V (b) from the graph V= 5 V when I = 5 A lost volts = 6 – 5 = 1 V V = Ir 1 = 5 x r r = 0·2 Ω

E is the emf, the p.d. across the battery when
no current flows in the circuit. R is the external resistance (load resistor) r is the internal resistance Terminal p.d. is the voltage across R Lost volts is the voltage across r I can be calculated several ways: I = lost volts ÷ r I = E ÷ (R + r) I = terminal p.d. ÷ R E r R

power matching

Electricity: Capacitors
A capacitor is a device designed to store charge. Circuit symbol

How does it store charge?
Capacitor in a d.c circuit electron arrives and repels another from the other side. - + - - +++ p.d. develops across the plates which opposes the p.d. of the battery. ----- +++++ when the p.d. across the plates equals the p.d. across the battery, the capacitor is fully charged

Experiment Aim: To find the relationship between the charge stored and the p.d. across a capacitor A B charge (nC) p.d. (V) V C The capacitor is charged when the switch is at A. The capacitor is discharged through the coulomb meter with the switch set to B.

charge (nC) p.d. (V) Q = constant V The ratio Q/V is called the capacitance. charge (C) Q capacitance (F) C = V p.d. (V)

Example Calculate the charge stored in a 2000 mF capacitor when the p.d. across the capacitor is 12 V. Q C = V Q 2000 x 10-6 = 12 Q = 0·024 C

The Energy Stored in a Capacitor Charge (C) p.d. (V) The energy stored in a capacitor is given by the area under the charge/voltage graph. Potential Difference (V) 1 Energy (J) E = Q V 2 Charge (C)

Alternative relationships for calculating the energy stored: Q 1 C = E = Q V 2 V 1 2 E = C V 2 Q 2 1 E = 2 C

Example The p.d. across a 2000 μF capacitor is 12 V. Calculate the energy stored in the capacitor. 1 E = C V 2 2 1 E = x 10-6 x 122 2 E = 0·144 J

Charging a Capacitor A V Current/ mA Time /s P.d. / V Time / s

Current (mA) P.d. (V) Battery p.d. V Imax = R Time (s) Time (s) When fully charged the current is zero. When fully charged p.d. is equal to the supply voltage.

Discharging a Capacitor A V Current/ mA Time /s P.d. / V Time / s

Current (mA) P.d. ( V) V Battery p.d. Imax = R Time (s) Time (s) When the capacitor is discharged the current and the voltage are zero.

Example A 2000mF V 12 V 1 kW Sketch graphs of I and p.d. against time as the capacitor charges. Calculate the final charge stored on the capacitor. Calculate the energy stored by the capacitor.

Current/A P.d./V 12 0·012 Time/s Time/s Q C = c) E = ½ QV = ½ 0·024 x 12 = 0·144 J b) V Q = CV = 2000 x 10-6 x 12 =0·024 C

Doping Electricity: Electrons at work Conductors, semiconductors and insulators Solids can be categorised into conductors, semiconductors or insulators by their ability to conduct electricity. Examples Conductors: copper, iron, aluminium Insulators : plastic, rubber, glass Semiconductors: silicon, germanium, selenium

Doping Electricity: Electrons at work Band Theory Band theory explains the different electrical properties of conductors, insulators and semiconductors. In an atom, electrons occupy discrete energy levels. When atoms combine to form a solid, the electrons become contained in energy bands separated by gaps. The two highest energy bands are called the valence band and the conduction band.

Doping Electricity: Electrons at work In an insulator the valence band is full. the conduction band has no electrons. the gap between these bands is large. there is no electrical conduction in an insulator conduction band band gap valence band

Doping Electricity: Electrons at work In a conductor the conduction band is not completely full. the conduction band contains electrons that are free to move conduction band valence band

Doping Electricity: Electrons at work In a semiconductor the gap between the bands is small. at room temperature there is sufficient energy for some electrons to move from the valence band to the conduction band. an increase in temperature increases the conductivity of the semiconductor. conduction band band gap valence band

Doping Electricity: Electrons at work Doping Adding an impurity to a semiconductor is called doping. N-type semiconductor silicon arsenic 4 outer electrons 5 outer electrons free electron

Doping Electricity: Electrons at work P-type semiconductor hole silicon 4 outer electrons indium 3 outer electrons

Doping Electricity: Electrons at work Doping reduces the resistance of a semiconductor. In n-type semiconductors conduction takes place by the movement of free electrons. In p-type semiconductors conduction takes place because electrons can move into the holes where an electron is missing, we say that conduction takes place due to the movement of positive holes.

Doping Electricity: Electrons at work The p-n Junction Diode p n p-type n-type

- + Doping Electricity: Electrons at work
potential barrier (0·7 volts) - + p n depletion layer – no free charge carriers

Doping Electricity: Electrons at work + - Forward-biased p n diode conducts Reverse-biased diode does not conduct - + p n

Doping Electricity: Electrons at work The Light Emitting Diode The LED has a p-n junction very close to the surface. When it is forward biased, electrons and holes cross the junction. Some of the electrons and holes meet and recombine giving out energy. The energy can be given out as a photon of light.

Doping Electricity: Electrons at work Band theory can explain the operation of an LED. n type p type junction conduction band conduction band valence band valence band When the LED is forward biased, electrons move across the junction from the conduction band in the n type to the p type. Electrons fall from the conduction band in the n type to the valence band releasing photons.

Doping Electricity: Electrons at work

Electricity: Electrons at work
Doping Electricity: Electrons at work LED Switch On Voltage supply voltage control empty LED socket blue LED green LED red LED switches for red, green, blue LEDs and empty socket voltmeter connector

Doping Electricity: Electrons at work LED Peak Switch On colour wavelength /nm voltage /V Infra Red Red Orange Yellow Green Blue Violet 760 660 600 580 540 480 430

Electricity: Electrons at work
Doping Electricity: Electrons at work switch on voltage / V 400 wavelength /nm

Doping Electricity: Electrons at work

Doping Electricity: Electrons at work Solar Cells Photons reaching the p-n junction have their energy absorbed, freeing electrons. A voltage is created by the separation of the electrons and holes. This is called the photovoltaic effect.

Higher Physics Electricity.

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