PH301 - ELECTRIC CIRCUITS
Motivation
Coverage
Understanding :
• Instruments and apparatus
• Communication
equipment
• Computational equipment
• Atoms and Molecules
• Electrical industry
•
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PH301
Electric Current
Resistance and Ohm’s Law
Sources of EMF
DC circuits
Kirchhoff’s Rules
Measurement
Transients
Alternating Current
1
Electric Current
Tipler chapter 25
• Defined as rate of flow • Thus
∆Q dQ
I
=
=
of electric charge
∆t dt
• Units are Ampere (A) • or Coulomb/second
• Conduction in wire.
• Free electrons accelerate,
collide, accelerate etc
• Average flow gives I
• ≡ Drift velocity
• Metals - electrons
• vdrift~10-4ms-1 cf vrms~ 106ms-1
• Ions in electrolytes
• Ionic Solutions - ions
PH301
2
Conduction
• Electron drifts at
• Switch on your
-4ms-1
v
~10
drift
bedroom light - how
long does it take for an time taken is3.6÷10-4 s
electron at the switch
=100hrs!
to reach the light bulb?
How can this be??
(suppose cable length
(Tipler
789)
of 3.6m)
PH301
3
Conduction
• dQ = qnAvdt
• n is number density
• q is particle charge
• v is drift velocity
• Current
dQ
=
dt
I = qnAvdrift
current density J=
v
•
I =qnv
A
PH301
Example: A particular length of copper
wire has a diameter of 1.02mm. It carries
current of 1.67A to a 200W light bulb.
What is a) the current density and b) the
electron drift velocity? (The density of
free electrons in copper is approximately
8.5x1028 electrons m-3.
4
Resistivity
• Charge q in E field experiences
a force, F, and results in I
• In many materials the current
density J ∝E (Ohmic conductors)
• The constant of proportionality
is the resistivity, ρ , (or its
inverse, conductivity) (obey
Ohm’s law)
• A wire of uniform cross-section
area A and length L, will carry
a current, I=JA. There is a
potential difference between its
ends V=EL.
PH301
F = qE
ρ = E/J
V
I
=ρ
L
A
or
V=
ρL
A
I
V = IR Ohm’s Law
R=
ρL
A
5
E M F and Circuits
A conductor will carry a steady
current if:
1. It is part of a closed loop i.e. a
completed circuit
2. The circuit contains an energy
source that drives the charge
• Electromotive force (EMF) can
be produced by chemical,
mechanical, thermal or
photoelectical means
• Sources of EMF have potential
difference between their (two)
terminals, V=E
• An Ideal Voltage Source has a
constant voltage at its terminals
regardless of the current drawn.
It has zero internal resistance.
•
+Eo R
• An Ideal Current Source
provides a constant current
regardless of its load.
It has infinite resistance
•
Io
PH301
6
Internal Resistance
• Real EMF sources are not
ideal
Can represent as either:
• Ideal voltage source Eo
with series resistance r
or
• Ideal current source I with
parallel resistance r
• PD at terminals V = Eo -Ir
or
V = I or
Eo =open circuit PD at terminals
r
Io
short circuit current
r
PH301
7
Energy and Power
• Power is rate of doing work • ∴ P = d(qV)/dt = Vdq/dt
=VI
dW/dt
•
P = I 2R
• Since V = IR (Ohm’s Law)
What is the power supplied by
a battery (non-ideal emf E with
internal resistance r) to a load
resistance?
Units: mks
Eo
r
R
• Energy in Joules J
• Power in Watts 1W = 1Js-1
• Potential Volts 1V = 1JC-1
PH301
8
Resistor Combination
PARALLEL
SERIES
V
I
R1
R2
R1
R1
R3
I
Current is same through each resistor
Equate sum of PD’s to total PD
V = V 1 + V 2 + V3 + …
= IR1 + IR2 + IR3 +…
= I(R1 + R2 + R3 +…) = IRe
Re = R1 + R2 + R3 + ...
R2
V
PD is same across each resistor
V = V 1 = V2
Total current divides eg
I = I 1 + I2
= V/ R1 + V/ R2 = V/Re
1/Re = 1/ R1 + 1/ R2 ….
PH301
9
Kirchhoff’s Rules
for solving more complex circuits
1)Algebraic sum of currents
into any junction is zero
ΣI=0
or current in = current out
2)Algebraic sum of PD’s around
any closed loop is zero
Σ emf’s = Σ IR
or sum of emf’s around circuit = sum
at any junction
≡ Charge conservation
Ex; determine the value of the currents
of potential drops across components
≡ Energy conservation
in each resistor of the following
circuit
R1=15Ω
E1=5V
R2=2 Ω R =8 Ω
3
E2=4V
PH301
10
Mesh Analysis
• A mesh is a loop containing no
smaller loops.
• Method automatically takes
account of second rule K2
• Allocate a current I1, I2 ..etc to
each mesh. Paths shared
between two adjacent mesh
loops are regarded as carrying
the algebraic sum from each
mesh. (see diagram)
a
b
I2
I1
d
e
g
current e d a b is I1
current b e is (I1 -I2) & f e is (I2-I3)
current f
h g
e is I3
c
I3
f
h
Try examples 26.17
in Tipler p808
PH301
11
DC Instruments
Ammeter
• Measures current
• Current passes through meter
• Ideally has zero resistance so
that circuit is unaffected
• Sensitive meters usually have
high resistance Rm (why?).
• Then can include small shunt
resistance in parallel Rs
current in meter Im = Rs/(Rm+Rs)I
Voltmeter
• Measures PD or voltage
• PD s appear across two points
• Can use an ammeter or
galvanometer in series with a
resistor Rs,,ideally infinite (why?)
• If max meter reading is Ifsd then
• Full scale voltage reading is
V= Ifsd(Rm + Rs)
V
A
PH301
12
Potential Divider
• The potential divider network
divides potential difference across
two resistances
• Potentiometer Circuit
to compare potential differences
current I = Vi/(R1+R2)
Vi
Vo = IR2 = ViR2/(R1+ R2)
potential is divided in the ratio of R’s
R1
R2
Vo
G
E1
E2
The circuit forms the basis of the
With E1connected, balance occurs when
potentiometer, used for the
E1= ViR2/(R1+ R2) Repeat with E2
measurement/comparison of emf.
instead : the ratio E1/E2 = R2(1)/R2(2),
The value of (R1+R2) is kept constant.
the ratio of the respective R2 values for
The galvo G detects NO CURRENT.
balance
PH301
13
Wheatstone’s Bridge
Comparison of Resistance
Each side of Bridge can be seen as a
potential divider.
Balance occurs when the potential at the
centre of each side is equal.
Va = Vb so that galvo current is zero.
R1
Vi
a
b
G
R3
Vi R1/(R1+R3) = Vi R2/(R2+R4)
R1/R3 = R2/R4
R2
R4
knowing ratio R2 /R4 , vary R3 to obtain
balance and hence measure unknown R1
PH301
14
Maximum Power Transfer
What is the condition that power
from a source V and resistance r
supplied to resistance R is a
maximum?
Power supplied P = I2R
current I = V/(R+r)
Show that
dP
dR
when
R= r
this is condition for maximum
power transfer to load R
V
r
Source
=0
R
Load
dP d 2
d
V2
=
I R=
dR dR
dR (R + r )2
V 2R
V2
= −2
+
=0
(R + r )3 (R + r )2
2R = R + r
R=r
PH301
15
RC circuits
time dependence of currents and voltages
•
•
Capacitor plates store charge +Q
and -Q so have overall neutrality
with PD of V between them
Q = CV defines Capacitance
(Coulomb,Farad,Volt)
R
+q
-q
•
•
i and q related : i = dq/dt
Rdq/dt = -q/C
so dq/q = -dt/RC
integrate ∫ lnq = ∫ -t/RC + constant
or use limits qo q and 0
C
Capacitor discharge
• Close switch at t=0, current i flows
until C is fully discharged
• Apply KII and definition of C
0= iR + q/C
t
or
q = q0 exp-t/RC
q = q0 at t=0 and decays
exponentially with time until Vc= 0 and
qo
q = 0 , RC is time constant
PH301
q(t)
RC
t
16
Charging a Capacitor
• Use same method as discharge:
R
q-CE = -CEexp-t/RC
q = CE(1 - exp-t/RC)
final charge at infinite t is qo=CE
i
+q
E
• Must put in limits, q=o at t=o
C
-q
obtain current by differentiation
i = dq/dt = (E/R) exp-t/RC = ioexp-t/RC
• after switch closed,a current io
flows immediately, decaying in
time. Apply KII and solve:
E = iR +q/C = Rdq/dt +q/C
So dq/dt = - (q-CE)/RC
or dq/(q-CE) = -dt/RC
[ ln(q-CE) = -t/RC ]tt=o
q(t)
E/R
i(t)
PH301
RC
t
17
Inductors and RL circuits
•
Switch Emf into RL circuit
See Tipler p939
• Inductors are usually coils of wire.
Current i will produce magnetic flux φ
and i ∝φ depending on geometry.
Self-inductance L defined by φ =L i.
Induced emf = -dφ/dt = -Ldi/dt
Voltage drop across inductor is Ldi/dt
voltage drop across capacitor is q/C
voltage drop across resistor is
iR
MEMORISE
RC has units of time …= time constant
L/R has units of time …= time constant
R
E
L
Determine current on switch closure t=0.
Use KII
E = Ldi/dt + Ri
Initially current i = 0 & di/dt=E/L
i
t
di
R
solve
= − dt
∫i−E/R
0
i (t ) =
[
∫L
0
E
1 − e −( R / L )t
R
]
nb i(0) = 0 and i(t) =E/R when t >> L/R
exercise: repeat above for switch shorting E
PH301
18
LC circuits
Discharge C through L
C
L
• A solution to this differential
equation is
q(t) = Qcos(ωt+φ )
(you can check this by back substitution)
d2q/dt2 = -ω 2Qcos(ωt+φ )
At t=0 i=0 and charge q on C is Qo
At time t (later) KII gives
Ldi/dt + q/C =0
and dq/dt=i so di/dt=d2q/dt2
L d2q/dt2 = -q/C
(1)
( compare this with SHM equation)
Solve for q(t) and hence i=dq/dt
(2)
Q is the maximum charge, ω the angular
frequency and φ the initial phase
comparing equations (1) and (2) gives:
frequency of oscillation ω = 1/√ LC
It is straightforward to determine the current
as previously indicated:
i=dq/dt = - ω Qsin(ωt+φ )
initial condition q =Qo and i = 0 at t=0
so q(t)= Qocosωt and i = - ω Qosinωt
PH301
19
• The charge on the capacitor
oscillates between +Qo and -Qo at a
(natural) frequency ω = 1/√ LC
• The current builds from zero to
ωQo (as C discharges) then reverses
sinusoidally to charge C in the
opposite direction
• Stored energy is transferred from
electrostatic (in C) to magnetic(in L)
and back in an oscillatory manner
• In an ideal LC circuit the oscillation
of charge and current continues
indefinitely.
• The equations are very similar to
those of Simple Harmonic Motion
RLC circuits
• normally resistance exists in the
circuit and absorbs energy.
• This causes the amplitude of
oscillation (both q and i) to decay
• The equation of motion is
obtained from application of KII
to the circuit and gives :
Ldi/dt + Ri + q/C =0
(3)
• Ld2i/dt2 + Rdi/dt + i/C =0 (4)
• This is the same form of equation
as that for Damped SHM
•
PH301
20
Damped Transient Oscillation
• The solutions to equations (3) and • When R2< 4L/C the circuit is
(4) will depend on the size of R
underdamped
• For small values of R ie R2‹ 4L/C • when R2= 4L/C the circuit
2
 1

settles fastest, without
R
−( R / 2 L )t


q (t ) = Qe
cos
t +φ
−
 LC 4 L2

oscillation, and is critically


damped.
• this is an exponentially decaying
2
sinusoidal oscillation (the circuit • When R > 4L/C the circuit is
overdamped and the solution is
is underdamped )
the sum of two decaying
• work out dq/dt and d2q/dt2 from
exponential terms
the equation above and substitute
into the differential equation (4) • See figure 31.14 and 31.15 in
Tipler
on slide 20 to verify that q(t) is in
fact a solution of equation (4)
PH301
21
Alternating Current
• Alternating current regularly reverses
and we normally consider sinusoidal
•
variation with time:
v(t)=vocos(ωt + φ1) or i(t)=iocos(ωt+ φ2)
vo and io are voltage and current amplitude
ω=2πf is frequency and φ is phase angle
• The phase difference between voltage
•
and current in the above example
would be (φ1 - φ2)
• Phase relationships are conveniently
•
represented on a phasor diagram
PH301
Phasors
The voltage and current are represented
as vectors, called phasors, rotating
anticlockwise in the xy-plane with
vo ω
angular frequency ω.
vo
ωt
ω
vocos(ωt)
i
ωt + φ1 o
ωt+ φ2
The observed voltage (or current) is then
the x-component or projection on the xaxis in the phasor diagram.
All phasors rotate at the same frequency.
22
Resistors in AC circuits
i
vocosωt
• Phasor diagram is therefore
io
R
ωt
• Apply KII to above circuit,
at any instant i = v/R so
i = vocos(ωt)/R = iocos (ωt)
vo
v=v0cos ωt
• Instantaneous power P =iv so
power P = iovocos2 (ωt)
= ½iovo(1 +cos(2ωt))
Current through and voltage across
R are always in phase
Peak voltage and current are
related by
vo = ioR
• Average power (over complete cycle)
PH301
‹P› = ½iovo = ½io2R
rms (root mean square) = io/√2, vo/√2
23
Inductors in AC circuits
• Phasor diagram:
i
vo
L
vocosωt
ωt
Apply KII
vocos(ωt) = L di/dt
i(t )
t
vo
integrating
∫ di = ∫ cos( ω t ) dt
i0
i=
L
ω
io
ω
• Inductive reactance XL = ωL
o
vo
v
π 

sin ω t = o cos  ω t − 
2
ωL
ωL

impedance to current (like a resistance)
there is a π/2 phase difference, the current
lags behind the voltage with amplitude vo/ωL.
• Power P(t)= iovosin(ωt)cos(ωt)
= ½ iovosin(2ωt)
• Average power is zero!!
PH301
24
Capacitors in AC circuits
i
vocosωt
+q
-q
• Phasor diagram
io
C
vo
ω
ωt
Voltage across C is vocos(ωt)=q/C
but dq/dt = i = - ωCvo sin(ωt)
= ωCvocos(ωt+/2)
Again there is a π/2 phase shift but
now the current leads the voltage
across the capacitor and has
amplitude v o
voltage
=
1 ω C reactance
• reactive impedance of capacitor
Xc = 1/ (ωC)
• Power =iv = iovosin(ωt)cos(ωt)
= ½ iovosin(2ωt)
• Average power is again zero
• Capacitors and Inductors on
average consume no power .
PH301
25
Series RLC with Generator
• Same current through all
components
I= IOcos(ωt)
• Calculate voltage across each
vR = VRcos (ωt)
VR= IOR
vL = VLcos (ωt + π/2) VL =IOXL
vC = VCcos (ωt - π/2) VC=IOXC
• Phasors add vectorially
vR+ vL + vC = v from K2
• thus V = VOcos (ωt + φ) and
voltages are related as on the
vector diagram as shown
• take projections onto x-axis
V is phase shifted by φ
ahead of current I (and VR)
R
L
V
C
VR
VL
IO
V
φ VR
ωt VL- VC
ωt
VC
PH301
26
V and I relations in AC
• Unit of Z is Ohm
• V2 = VR2 + (VL - VC)2
•
= (IOR) 2 + (IOXL - IOXC)2
= IO2[R2 + (XL - XC)2]
•
VO = IOZ with impedance of
the circuit Z = √ [R2 +(XL-XC)2]
• Impedance in AC plays the
same role as resistance in DC
note: it depends on R,L C and ω
For AC series circuits add
resistances and square, then add
inductive impedance(reactance)
subtracting capacitative
impedance(reactance) and square
reactance term before adding to
squared resistance.
• Phase angle of V given by
V − VC I O ( X L − X C )
tan φ = L
=
VR
IO R
PH301
 ωL − 1 / ωC 

R


φ = tan −1 
27
AC Impedances
• Resistor v = iR
• Inductor vo = ioωL
• Capacitor vo = io/(1/ωC)
• Series combination:
Z=√
[R2
v in phase with i
v is π/2 ahead of i
v is π/2 behind i.
 ωL − 1 / ωC 
φ = tan 

R


−1
+(XL-XC)2]
2
1 

Z = R 2 +  ωL −

ωC 

PH301
28
Series Resonance
• ω0 is the resonance frequency
V
IO = O
Z
Z =
IO
R 2 + ( X L − X C )2
• Resonance occurs when current
and voltage from emf are in
phase.
• φ = 0 when ωL = 1/ωC
• Z is then a minimum Z=R and
circuit is purely resistive
ω = ω0 =
1
LC
Increasing R
∆ω
ω0
ω
• Bandwidth depends on
Q = ωoL/R = ω0/∆ω
PH301
ω
29
LCR Parallel Resonance
• Voltage now same across each
component v = Vocos(ωt)
Parallel circuit and resonance
i
v
iC
C
iR
R
L
iL
• Magnitude and phase differs in
each component
VO
cos( ω t )
R
π

iC = ω CV O cos  ω t + 
2

VO
π

iL =
cos  ω t + 
ωL
2

iR =
Phasor diagram
IO
Vo
IC
IC-IL
φ
IR
ωt
IL
2
V 
IO = IR2 +(IL − IC)2 =  
 R
PH301
2
V V  V
+ −  =
 XL XC  Z
30
LCR Parallel Resonance ctd.
• Impedance is given by
1
=
Z
2
1  1

C
+
−
ω
  

R
L
ω
  

• The phase angle between current
and voltage at the generator is
given by
I C − I L ωC − 1 / ωL
tan δ =
=
IR
1/ R
2
• At resonance
•
ωo = 1/√LC and
Z has its max value Z=R
This is because, at resonance XC=XL ,
and the currents in the inductor and
capacitor are equal but 180o out of
phase (cancel each other out).The
net current is that in the resistor
PH301•
Power in the tuned circuit is
consumed only in the resistor.
Average P=IR2R = IO2 Rcos2δ
or
P=IOVORcos δ
cf for series tuned circuit average
P= IR2R = IO2 R
or
P=IOVORcos δ
31
cos δ is called the Power Factor
Thevenin’s Theorem
•
•
•
Linear network with two accessible
terminals can be replaced by emf in
series with impedance.
Linear
circuit
Emf E is open circuit PD between
terminals.
E
≡
Impedance is that presented by
system to the terminals when all
internal emf’s have been replaced
by resistances equal to their
internal resistance.
R
E
PH301
When connected
to other circuit
behaves as simple
voltage generator
in series with R
32
Mutual Inductance: Transformer
•
•
•
•
•
Transformer is magnetically
coupled pair of coils, selfinductance L1 and L2.
Mutual inductance produces output
V2 from input V1.
Flux in L1 is N1φ and in L2 is N2φ
where φ is flux in core. Assume all
flux from 1 goes through 2 and
vice-versa.
V1 = N1dφ /dt and V2= N2dφ /dt
V2= V1N2/N1= V1n
Connect load RL ,draws IL and
N1I1= -N2I2 (why? See Tipler 958)
• No losses: V1I1= V2I2
• Applications of transformer
n=N2/N1
V1
N1
N2
V2
RLoad
• Power transmission
• Signal inversion
• Matching, to achieve max power
transfer
Power in load = (I2)2RL = (V2)2/RL =(V1n)2/RL
Power as seen from input=(V1)2/Rreflected
Rreflected= RL /n2
PH301
33