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Announcements
• First lab this afternoon SHL 016
– Introduction to EWB
http://www.physics.udel.edu/~jholder/Phys645/index.htm
Tuesday:
Matthew Bihler
Gaurav Pandey
Philip Zandona
Trisha
Wednesday:
Joseph Brosch
Erin Grace
Ali Jafri
Lei Chen
Halise Celik
Fatih
Jingliang Zhang
Eric
Lecture 2 Overview
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Kirchoff’s Laws
Power and Resistance
Practical sources
Voltage/ Current Dividers
DC circuit analysis
– Element combination
– Using Kirchoff’s Voltage Law
– Using Kirchoff’s Current Law
– Mesh Analysis
Kirchhoff's three laws of spectroscopy
1. A hot solid object produces light with a continuous
spectrum.
2. A hot tenuous gas produces light with spectral lines
at discrete wavelengths (i.e. specific colors) which
depend on the energy levels of the atoms in the gas.
(emission spectrum)
3. A hot solid object surrounded by a cool tenuous
gas (i.e. cooler than the hot object) produces light with
an almost continuous spectrum which has gaps at
discrete wavelengths depending on the energy levels of
the atoms in the gas. (absorption spectrum)
Gustav Robert Kirchhoff
(12 March 1824 – 17 October 1887)
Kirchoff’s Current Law
The sum of the current at
any node must equal
zero: i.e., the current
flowing into a node must
equal the current flowing
out of the node
Conservation of Charge
(Current = rate of change of charge)
At Node 1:
-i+i1+i2+i3=0
i=i1+i2+i3
Note convention: current flows from positive terminal
In order for current to flow, there must exist a closed circuit
Kirchoff’s Voltage Law
The sum of the voltages
around a closed loop is
zero
Potential: at a=va
b=vb
Potential difference:
v2=va-vb
Around the loop:
-v1+v2=0
v1=v2
Note: potential measured relative to ground: true ground (earth)
or chassis ground (enclosure)
Circuit elements and their i-v characteristics:
Resistor
Ohm’s law: Current is proportional to applied Voltage,
and inversely proportional to the Resistance: V=IR
v
L
m
R    .m  2  
A
m
1


R = resistance: depends on materials and
geometry
ρ = resistivity: depends only on materials
σ = conductivity
Resistor Color code
http://www.dannyg.com/examples/res2/resistor.htm
560KΩ ± 5%
Electric Power
Electric power = amount of work done/unit time
Voltage V = work/unit charge
So, to move charge Q, work done=VQ
Power = VQ/t =VI
P=IV =I2R =V2/R
Units = Joules/sec = Watts (W)
What do we pay for in the electricity bill?
kWh=energy/time × time = energy
More on Power and Resistance
What do the power ratings of appliances mean?
e.g. what does a 1000W hair dryer tell us?
• Assume 120V (USA) if voltage not specified.
P 1000

 8.33A
Max current
V 120
V2
V 2 1202
P
R

 14.4
R
P 1000
P  iV  i 
Never exceed the rated power
Can you use this appliance on a 240V (UK) line?
V 2 2402
P

 4000W
R 14.4
Destroys the appliance!
Resistance Limits:
Open and Short Circuits
• Short Circuit: A wire! R=0, V=0 for any i.
– Particularly bad for any voltage source
• Open circuit: A break! R→, i=0 for any V.
– Particularly bad for current source
Series Resistors and the Voltage
Divider Rule
For N resistors in series:
N
N
n 1
n 1
REQ   Rn , i  i1  i2  ...  iN , v   vn
REQ>(R1,R2,…..,RN)
Voltage Divider:
R1
v1 
(1.5V)
REQ
R
v2  2 (1.5V)
REQ
v3 
R3
(1.5V)
REQ
R3
Practical Voltage Sources
Modelled with an ideal source and a
series resistor
Ideal voltage source: rS=0
Current is the same at all points
vS
vS
iS 

REQ rS  RL
RL
vL  iS RL 
vS
rS  RL
voltage divider
So for a practical voltage source, the
output voltage depends upon RL
If rS<< RL , vL = vS , independent of RL
Parallel Resistors and the Current Divider Rule
For N resistors in parallel:
N
N
1
1
  , v  v1  .....  vN , iS   in
REQ n 1 R n
n 1
REQ<(R1,R2,…..,RN)
v REQ
iS
Current Divider: i1  
R1
R1
REQ
v
i2 

iS
R2
R2
v REQ
i3 

iS
R3
R3
Large current through smaller R.
Advantage of a parallel circuit; a broken branch will not affect
other branches
Practical Current Sources
Modelled with an ideal source and a
parallel resistor
Ideal current source: rS= 
iL 
REQ 
1
REQ
RL
iS
current divider
rS RL

 1 1  rS  RL
 

 rS RL 
iL 
rS
iS
rS  RL
The output current now depends upon RL
If rS>> RL , iL = iS , independent of RL
Circuit analysis method 1:
Apply element combination rules
Series resistors
Parallel resistors
Series voltage sources
Parallel current sources
Circuit analysis method 1: element
combination
R=22Ω
R1=10Ω
R2=20Ω
R3=30Ω
V=10V
Find the equivalent resistance and the current at I
I= 10V/22Ω = 0.45A
Circuit analysis method 2a: KVL and KCL
Kirchoff’s Voltage Law
The sum of the voltages around a closed loop must be zero
• Draw the current direction (arbitrary)
and label the voltage directions
(determined by the defined current
direction). Voltage on a voltage source is
always from positive to negative end.
• Define either clockwise or counterclockwise as positive direction for
summing voltages. Once the direction is
defined, use the same convention in every
loop.Voltage across a resistor is +’ve if
voltage direction the same as current
direction, -’ve otherwise
• Apply KVL
Kirchoff’s Voltage Law: Multiloop
The sum of the voltages around a closed loop must be zero
• Draw the current direction (arbitrary)
and label the voltage directions
(determined by the defined current
direction).
R3
• Define either clockwise or counterclockwise as positive voltage direction.
Once the direction is defined, use the
same convention in every loop.
• Apply KVL
   Vr  V1  V2  0
 V2  V3  0
  Ir  IR1  I 2 R2
 I 2 R2  I 3 R3  0
Kirchoff’s Current Law
The sum of the current at a node must be zero: Iin=Iout
I=I2+I3
(1)
ε=Ir+IR1+I2R2 (2)
I3R3-I2R2 =0 (3)
R3
Set r=1Ω, R1=3Ω, R2=5Ω, R3=10Ω, ε=3V
I- I2- I3 = 0
(4)
4I+5I2+ 0I3= 3
(5)
0I- 5I2+10I3= 0 (6)
Last note on KCL KVL analysis
• If solutions to currents or voltages are
negative, this means the real direction is
opposite to what you originally defined
• To deal with current sources: current is
known, but assign a voltage across it
which has to be solved
Sample Problem