Electricity – Current, potential difference, power and resistance.
2.1 ResistanceWhen there is a current in a circuit, there are collisions between the ions making up the conductor and the electrons (charge carriers). These collisions result in the motion of the electrons being impeded. The amount of charge flowing in one second is reduced and the conductor is said to have resistance. The more difficult it is for the electrons to get through (the more collisions) the bigger the resistance of the conductor.
Also as a result of the electron-ion collisions, the ions are forced to vibrate more violently. This increases their average kinetic energy and hence the temperature of the conductor (from kinetic particle theory) i.e. heat is produced.
Resistance is the opposition to current in a conductor.
Factors affecting the resistance of a conductor are:
(a) the longer the conductor is, the greater it’s resistance i.e. R α l;
(b) the thinner (smaller cross-sectional area) of the conductor, the greater it’s resistance; (c) the resistance is dependent on the material used;
(d) the resistance is dependent on the temperature of the conductor.
Rheostats (variable resistors)
This is a device used to control the current in a circuit. It does this by changing the length of wire in the circuit and so altering the resistance.
Ohm’s Law
The voltage across the ends of a conductor is directly proportional to the current in the conductor provided the physical conditions (e.g. temperature) are unchanged.
i.e. V = constant I
This constant is called the resistance, R, of the conductor
Thus V = IR R = V I = V
I R
volts (V) amperes (A) ohms ()
Voltage across a resistor = current in the resistor x resistance of resistor I / A
0 V / V
V α I
2.2 Power and Energy Dissipated in a Resistor
Power = energy transferred in one second Power = work done in one second
Electrical power = electrical energy transferred time taken for transfer Electrical power = charge transferred x voltage
time taken for transfer
Electrical power = Q x V but current I = charge transferred = Q
t time taken t
Electrical power = IV but V = IR and I = V
R Hence P = IV = I x (IR) = I2R P = IV = V x V = V2 R R i.e. P = E = Wd = QV = It V = IV = I2R = V2 t t t t R
Electrical energy = Electrical power x time
E = Pt = IVt = I2Rt = V2t R I V I R
CfE Higher Physics – Electricity – Current, potential difference, power and resitance. Page 3 2.3 Voltages of Sources
The electromotive force (e.m.f.) is the energy given to 1 coulomb of charge by the source of electrical energy
A 12 V battery will give each coulomb of charge passing through it 12 J of energy – this energy will be dissipated as heat in the resistors as charge flows round the circuit.
e.m.f. of battery = IRAB + IRBC + IRCD
2.4 Resistors in Series
≡
RS is the effective series resistance which produces the same effect as resistors R1, R2 and
R3 connected in series i.e. the same current I in both circuits and the same voltage V
across AD.
Applying the Conservation of Energy for one coulomb of charge from A to D gives: energy transferred in moving 1 C = energy transferred in moving 1 C from
from A to D A to B plus B to C plus C to D
V = V1 + V2 + V3
IRS = IR1 + IR2 + IR3
RS = R1 + R2 + R3
What you need: resistors, ohmmeter.
What to do: Measure the resistance of the three individual resistors. Connect these three resistors in series and measure their combined resistance i.e. RS.
Resistor Resistance /
R1
R2
R3
RS
Write down the expression for the effective resistance of a number of resistors in series?
I RS V A D − + I I I R1 R2 R3 V V3 3 A C V2 3 V2 3 B D − + + emf = 12 V 2 Ω 1 C (12 J) 1 C (12 J) 1 C (6 J) 1 C (6 J) 1 C (4 J) 1 C (4 J) 1 C (0 J) 1 C (0 J) 6 Ω 4 Ω p.d = 4 V p.d = 6 V p.d = 2 V A B D C
2.5 Resistors in Parallel
≡
RP is the effective parallel resistance, which produces the same effect as resistors R1, R2
and R3 connected in parallel i.e. there is the same current I from X to Y and the same
voltage V across XY. In both circuits applying the Conservation of Charge to the charge transferred round the circuit in one second (i.e. current) gives:
charge transferred in 1 s through RP = charge transferred in 1 s through R1plus R2 plus R3
I = I1 + I2 + I3
V = V + V + V RP R1 R2 R3
1 = 1 + 1 + 1
RP R1 R2 R3
What you need: resistors, ohmmeter.
What to do: Measure the resistance of the three individual resistors. Connect the three resistors in parallel and measure their combined resistance i.e. RP.
Resistor Resistance / Ω 1 . Resistance R1 R2 R3 RP
How does the resistance of the combination compare with the value of the individual resistors?
Write down the expression for the effective resistance of a number of resistors in parallel.
Note: If two identical resistors are connected in parallel then the effective resistance of the pair is half of one of the resistors.
I2 I1 I3 I I R1 R2 R3 V − + V X Y I I RP V − + V X Y
CfE Higher Physics – Electricity – Current, potential difference, power and resitance. Page 5 2.6 Potential Dividers
A potential divider is an electrical circuit that enables different voltages to be taken from one source.
E = V1 + V2 assuming source has
negligible internal resistance. If R1 = R2 then V1 = V2
If R1 > R2 then V1 > V2
If R1 < R2 then V1 < V2
Circuit current = I = voltage of source = E .
total resistance R1 + R2 V1 = IR1 = E x R1 . R1 + R2 V2 = IR2 = E x R2 = VXY R1 + R2 Also I = V1 = V2 R1 R2
V1 = R1 i.e. the voltage ratio is equal to the resistance ratio
V2 R2
R1 and R2 could be replaced by a variable
resistor as shown so that any required
voltage between 0 and E volts can be tapped off between X and Y.
R1 R2 + E − V2 V1 X Y + E − V X Y
2.7 The Wheatstone Network
If the value of the resistors are adjusted so that there is no reading on the voltmeter (i.e. Null deflection) then the network is said to be balanced.
The points B and C are thus at the same potential. Current in RX is equal to the current in
R1; current in R2 is equal to the current in R3.
Thus p.d. across RX = p.d. across R2
p.d. across R1 = p.d. across R3
Therefore p.d. across RX = p.d. across R2
p.d. across R1 p.d. across R3
I1RX = I2R2
I1R1 I2R3
Hence RX = R2 or RX = R1
R1 R3 R2 R3
If the values of three of the resistors R1, R2 and R3 are known, the value of the fourth RX
can be found. Normally the ratio of R1 and R3 is known i.e. R1/R3 (a simple ratio e.g.
1:10).
The resistance box value, R2 is adjusted until the network is balanced i.e. no reading on V.
This is a very accurate method to measure resistance provided V is a sensitive voltmeter.
The conditions for balance are unaffected by changing the positions of the voltmeter and the cell.
Although this method may be used over a wide range of resistance values, if RX is very
small, the resistance of the leads will cause appreciable error. By making the ratio of
R1/R3 small e.g. 1:1000 then the value of R2 will have to be large to give balance and so
the accuracy is to 4 significant figures:
RX = R1 i.e. RX = 1 x R2 R2 R3 1000 A + − R3 V R2 R1 RX D B C R1 RX A D R3 R2 B V C R1 RX A D R3 R2 B V I1 I2 C
CfE Higher Physics – Electricity – Current, potential difference, power and resitance. Page 7 What you need: Resistance box; two resistors in ratio 1 to 10; unknown resistors; voltmeter; ohmmeter.
What to do:
1. Set up the circuit and adjust the value of Runtil the network is balanced (reading on V is zero).
2. Calculate the resistance of the unknown resistor, RX using RX = 1 – enter this value in
the table. R 10
3. Remove the unknown resistor from the circuit and measure it’s resistance with the ohmmeter – enter this value in the table.
4. Use the colour code to obtain the value of the unknown resistor – enter this value in the table.
5. Repeat steps 1 - 3 for other unknown resistors.
Wheatstone network value of RX / Ω
Ohmmeter value of RX / Ω
Resistor code value for RX / Ω Comment on these results.
1 RX 10 R V resistance box
2.8 Wheatstone Network in Out of Balance Conditions
What you need: Resistance box; 1:10 and 1 k resistors; voltmeter.
What to do:
1. Set up the circuit and adjust the value of RXuntil the network is balanced (reading on V
is zero).
RX = 1
R 10
2. Increase the value of Rx in stages of 5 and note the reading on V in the table. 3. Re-balance the network.
4. Decrease the value of Rx in stages of 5 and note the reading on V in the table. Reading on V (V) Change in Rx () 20 15 10 5 0 –5 –10 –15 –20
5. Draw a graph of how the reading on V varies with change in Rx – what do you notice? 1 RX 10 R = 1 kΩ V resistance box RX / Ω V / V 0
CfE Higher Physics – Electricity – Current, potential difference, power and resitance. Page 9 A Useful Circuit
The fact that a small change in resistance in one arm of a Wheatstone network gives a proportional change in voltage is very useful because a resistance change can be made in several ways.
The variable resistor, in each circuit, is used to balance the bridge circuit at the start.
The strain gauge is fixed on the wall across the crack. If the crack widens the gauge will be twisted and its resistance changes. The reading on V will change so that
movement in a building can be checked.
The component R will change its resistance if its temperature changes. The reading on the meter is, therefore, a measurement of the temperature. A useful thing about this circuit is that the meter showing the
temperature can be a long way from the heat e.g. a furnace temperature can be shown in the control room.
The light dependent resistor will change its resistance as the intensity of the light falling on it changes. The meter reading is a measure of light intensity.
