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Internal Resistance, EMF and Oscilloscopes
What we are going to achieve today
• Creating power from lemons• Find out about internal resistance of batteries• Know what is the electromotive force is• Use an oscilloscope to measure the frequency
and voltage from an signal generator
Using lemons, limes and potatoes to power an LED
zinc copper
Making Batteries
There is nothing special about batteries – but these have a high internal resistance.
Rl
What do you think the best way to minimise the internal resistance of your battery?Think about resistivity (lemon is a poor conductor)
Batteries
What is the main energy transfer in a battery?
Electrical energy
This is powered by two 1.5 V AA cells in series - what is the supply voltage?
What is the emf of the supply?
After a while the battery needs to be replaced; why?
What determines how quickly it runs down?
What determines how much current is drawn from the supply?
EMF Electromotive Force
• EMF is the external work expended per unit of charge to produce an electric potential
difference across two open-circuited terminals
This is the same definition as voltage but on an open circuit (no current flow)
Why is this definition important?
Batteries have internal resistance
The circuit now has two resistors in
The internal resistance of the battery, r is very small.
R is much larger
The total resistance of this circuit is Rtotal = R +r
I = E / (R+r)
Batteries have internal resistance
As charge goes around the circuit the sum of emfs must equal the sum of voltage drops leading to
EMF = I R + I r
The terminal voltage is equal to I R so this can be rearranged to give:
V = E – I r
and interpreted as terminal voltage = emf – ‘lost volts’
R -small
Rtotal = r + R -small
The current will now be larger as the total resistance of the circuit is much lower
The voltage lost across r
V = I(large)r
The voltage lost will now be a problem
This case is of a small load resistance connected to a battery is seen in a starter motor on a car
http://www.youtube.com/watch?v=al6Yz3Nv7dY
http://www.youtube.com/watch?v=Ut7yBdIehYY&NR=1
Starter motor on a conventional carThe headlamps are connected in parallel across a twelve-volt battery.
The starter motor is also in parallel controlled by the ignition switch.
Since the starter motor has a very low resistance it demands a very high current (say 60 A).
The battery itself has a low internal resistance (say 0.01 Ω).
The headlamps themselves draw a much lower current (they have a higher resistance)
Lamp
R
Starter motor
Ignition switch
12V
What will happen to the lights?
A use of high internal resistance
Quick Questions
1. A 9.0 V battery has an internal resistance of 12.0 . (a) What is the potential difference across its terminals when it is supplying a
current of 50.0 mA?
(b)What is the maximum current this battery could supply?
(c)Draw a sketch graph to show how the terminal potential difference varies with the current supplied if the internal resistance remains constant. How could the internal resistance be obtained from the graph?
2. A cell in a deaf aid supplies a current of 25.0 mA through a resistance of 400 . When the wearer turns up the volume, the resistance is changed to 100 and the current rises to 60 mA. What is the emf and internal resistance of the cell?
V = E – I r V = IRE = I(R +r)
You need to set up a simultaneous equation
Answers1. (a) pd = E – I r = 9 – (50 x 10-3 x 12) = 8.4 V (b) Max current = E/r = 9 / 12 = 0.75 A
2. E = I(R +r)E = 25 x 10-3 (400 + r) and E = 60 x 10-3 (100 +
r)So25 x 10-3 (400 + r) = 60 x 10-3 (100 + r) so r =
114.3 E = 10 + (25 x 10-3 x 114.3) = 12.86 V
I
V
OscilloscopeUse the signal generator to create A/C (alternating current) on the Oscilloscope
Draw 2 signals of 2 different frequencies
Work out the frequency of the Oscilloscope
Add the magnitude of the waveheight
How do you use an oscilloscope as a dc and ac voltmeter?How do you use it to measure frequency?
Voltage
what the average value of an ac voltage or current is over a whole number of cycles?
power is calculated by P = IV, both I and V change sign together, so power is always positive
Positive power
calling the maximum value of I p Vp and a minimum of zero
0V
P = I p Vp
Average power = ½ I p Vp
P = 1/2 I pV p = 1/2 I p 2 R = 1/2 V p 2 / R (using V = I R)
RMS voltage (root mean squared)RMS links the dc equivalent values to ac peak values.
The point is that a sinusoidal ac supply of peak value Vp delivers the same average power as steady dc of value Vp / √2
The same power would be delivered by a dc with values I and V if: P = I2R= 1/2 I p 2 R and P = V 2 / R = 1/2 Vp 2 / R I2R= 1/2 I p 2 R (divide both sides by R) V 2 / R = 1/2 Vp 2 / R
(multiply by R)These lead to the equations: I2 = I p 2 / 2 V 2 = V p 2 / 2
(square root)I = Ip / √2 V = Vp / √2
RMS Graph
listed voltages for power outlets, e.g. 120 V (USA) or 230 V (Europe), are almost always quoted in RMS values, and not peak values
Questions1. The mains ac supply in some countries is 110 V r.m.s at 50 Hz (sinusoidal).(a) What is the peak value of voltage?(b) What is the peak-to-peak value of voltage?(c) How long does one cycle of this ac supply last?(d) A 100 W lamp is designed for use with 110 V ac What is the resistance of its filament? (use r.m.s. value for V)
2. Which will light a lamp more brightly, 12 V peak ac or 12 V steady dc? Explain.
3.UK mains ac has an r.m.s. value of 230 V and a frequency of 50 Hz. Sketch a graph of voltage against time for one cycle of this ac and include values for peak voltage and time period on the axes.
4. What is the ratio of powers delivered by 20 V dc and 20 V peak ac to the same load?
Vp =r.m.s x √ 2 Wavelength = 1 / frequency
Answers and Worked Solutions1. (a) 110 x √ 2 = 155.6 V (b) 2 x 155.6 = 311 V (c) 1 / 50 Hz = 0.02s (d) P = V2 /R R = 12100/100 = 121
2. 12 V peak ac is equivalent to 12/√ 2 V = 8.5 V dc equivalent. So 12 V peak to peak is dimmer.
3. 230 x √ 2 V = 325V, so graph varies between + 325 V and -325 V.
One time period = 1 / 50 Hz = 0.02s
4. P = V2 /R so the ratio of powers is ratio of voltages squared. 20 V peak ac is equivalent to 20/ √ 2 V dc i.e. 14.14 V so dc power / ac power = 202 /14.142 = 2
or dc V2 / (ac peak / √ 2)2 = (√ 2)2 =2