Transcript DC circuit theory

DC circuit theory
Learning outcomes
• explain the behaviour of DC circuits using concepts of EMF,
internal resistance of power sources and potential dividers
• give a microscopic description of resistance in a wire
• define and use concepts of resistivity and conductance
• state Kirchhoff’s laws and use them to analyse DC circuits
• define capacitance and solve DC circuit problems involving
capacitors, including energy stored
• carry out related practical work (using voltmeter, ammeter,
multimeter, micrometer)
Teaching challenges
It is always advisable to revisit concepts introduced at
KS3 and GCSE level, to identify misconceptions
about electricity and (try to) correct them.
In pairs:
Make a spidergram showing key concepts related to
electric circuits, and relationships between them.
EMF and potential difference
A-level: A battery maintains an electric field through the circuit.
This enables it to do work on charges wherever there is a
potential difference e.g. in a filament.
Electromotive force is the energy supplied per unit
charge. (work done on each coulomb of charge)
Potential difference (p.d.) is the energy transferred per
unit charge between 2 points in a circuit.
(work done by each coulomb of charge)
Unit (for both) is the volt = joule/coulomb
Resistor networks
Resistors in series
V = V1+ V2 [conservation of energy]
IR = IR1 + IR2
R = R1 + R2
R is always larger than any of R1, R2 etc
Resistors in parallel
I = I1 + I2 [conservation of charge]
V/R = V/R1 + V/R2
1/R = 1/R1 + 1/R2
R is always smaller than any of R1, R2 etc
Potential dividers
V1 IR1 R1


V2 IR2 R2
Useful for constructing sensors
In pairs, sketch
• a dark sensor
• a heat sensor
• a cold sensor
Real power supplies
Demonstrations:
• 12V DC supply lighting more and more lamps in parallel
• EHT with a 1.5V lamp
What’s happening?
E = Vinternal ('lost volts') +Vexternal
E = I(r + R)
IR = E - Ir
terminal V = E - Ir
Graphical representation
y = mx + c
V = -rI + E
Resistance in a wire
microscopic picture: free electrons drifting through a metal
(polycrystalline, each crystal having an ionic lattice)
Rl
1
R
A
l
R
A
RA

l
• constant of proportionality is resistivity, unit  m
• a material property
Compare with rules for R networks. VPL simulation.
Current and charge
Current is rate of flow of charge
Q
I
t
e.g. 1016 electrons pass a point every second
10 1.6 10 C
I
 0.0016C  1.6 mA
1s
Demonstration: Conduction by ‘coloured’ ions
16
19
Drift velocity
Q  number of charged particles charge per particle
Q  (nAx)e
where
• n is the number of free electrons per unit volume
• A is the cross sectional area
• Δx is a small length along the wire
• e is the charge of an electron
DQ nADxe
=
Dt
Dt
Dx
vd =
Dt
I = nAvd e
I=
Comparing copper with tungsten
metal
copper
electrons electron drift velocity in
per m3
mm s-1
8.5 ×
1028
~0.02
~250
3.4 ×
1028
The difference in drift velocities explains why
tungsten
incandescent lamps glow white hot while their
connecting wires stay safely at room temperature.
Conductivity
Metal wires conduct extremely well.
Conductance G = I / V , unit siemens (symbol S)
• depends on the number of carriers available
• ratio I / V is 'effect per unit of cause‘
Note:
• conductance is the reciprocal of resistance
• conductivity,  [unit S m-1] is the reciprocal of resistivity
Capacitance
• a measure of how much charge a
capacitor can separate at a given p.d.
• unit of capacitance: farad (symbol F)
charge
capacitance 
p.d.
Q
C
V
1
1
Energy stored,W  QV  CV
2
2
demonstration super-capacitor
Note: There are rules for adding capacitors in networks.
2
Lab practicals
• internal resistance of a potato cell
• resistivity of a wire (using micrometer)
• charging and discharging a capacitor
Kirchhoff’s 1st law
The total current
entering a circuit
junction equals the
total current
leaving it.
[conservation of charge]
Kirchhoff’s 2nd law
The sum of the emfs round a loop in any circuit
= the sum of the p.d.s round the loop.
[conservation of energy]
E1 + E2 + E3 + … = I1R1 + I2R2 + I3R3 + …
where I1, I2, I3 … represent currents through the resistances R1, R2,
R3 …
Physlets (simulations): ‘Second semester’< ‘DC Circuits’
• ‘Kirchhoff's Loop Rule’
• ‘Applying Kirchhoff's Rules’
Kirchhoff’s 2nd law - example
A circuit consists of a cell of
emf 1.6 V in series with a
resistance 2.0  connected to a
resistor of resistance 3.0  in
parallel with a resistor of
resistance 6.0 .
Determine the total current
drawn from the cell and the
potential difference across the 3.0
 resistor.
Solution
Consider the circuit loop consisting of the cell and
the 3.0  resistor:
1.6 V = 3 I1 + 2 (I1 + I2)
Thus 1.6 V = 5 I1 + 2 I2 …(1)
Consider the circuit loop consisting of the cell and
the 6.0  resistor:
1.6 V = 6 I2 + 2 (I1 + I2)
Thus 1.6 V = 2 I1 + 8  I2 …(2)
Subtracting the second equation from the first
gives:
0 V = 3 I1 - 6 I2
hence I1 = 2 I2
Substituting I1 = 2 I2 into the second equation
gives:
1.6 V = 12 I2
Thus I2 = 0.13 A and I1 = 0.27 A
Current through cell = I1 + I2 = 0.40 A
pd across 3.0  resistor = I1 × 3.0
(= I2 × 6.0 ) = 0.8 V
Endpoints
Related topics
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sensors of many types use the potential divider principle
factors affecting capacitance (plate spacing & area, dielectric material)
exponential nature of charging and discharging capacitors
how ammeters and voltmeters affect circuit behaviour
maximum power theorem
AC circuit theory