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PHYS219 Fall semester 2014
Lecture 08: Circuits, Fuses and
Kirchhoff’s Laws
Dimitrios Giannios
Purdue University
The Idea of a Circuit Diagram
wire
Load
emf
wire
battery
wire
wire
Assumption:
Negligible voltage
drop across wire
Circuit Symbols
Open Circuits
continuous path
open path
Power Dissipation - the Technology Underlying
Fuses and Circuit Breakers
•In a fuse, current passes through a thin metal strip
– The strip acts as a resistor with a small resistance
– If a fault causes the current to become large, the increased power melts the strip to melt and the
current stops – open circuit!
•A circuit breaker “OPENS” when current exceeds a
predetermined limit
– The circuit breaker can be reset
dc Circuit Analysis
• An electric circuit is a combination of connected elements forming a
continuous path through which charge is able to move
• Calculating the current in a circuit is called circuit analysis
• dc stands for direct current - the current is constant
• The current is viewed as the motion of the positive charges traveling
through the circuit
• Exact values of current and voltage depend on where in the circuit you
are measuring
Kirchoff’s Loop Rule
The change in electrical potential of a charge q
as it moves around a closed loop MUST be zero
q
Recover Ohm’s Law:
Power Dissipated in Resistors
I
q
Energy Decrease = qΔV
q = I δt
Energy Decrease = (Iδt)ΔV = Energy Dissipated in R
Power dissipated = Energy/δt = IΔV = IV
using Ohm's Law : V = IR
Power dissipated = IV = I2R = V2/R
units : [Joule / sec] = [Watt] = [W]
How Much Power is
Dissipated in the Load?
How Much Power is Dissipated in
the Resistor?
9.0 V
360 mA
Resistors in Series
Change in potential
around closed loop
= + 𝜀 – IR1-IR2 = + 𝜀 – I(R1+R2)
= +𝜀 – IRequiv = 0
junction or node
Resistors in Parallel
Kirchoff’s Junction Rule
I1+I2=I
The currents are NOT independent!
1/Requiv=1/R1 +1/R2
ε = I Requiv
how can we calculate I1 and I2 ?
ε - I1 R1 = 0, ε - I2 R2 =0
I = I1 +I2
junction or node
Resistors in Parallel (more)
I1+I2=I
More complicated circuits
I
a)
10Ω
9 V
50Ω
i1
i2
30Ω
b)
10Ω
9 V
18.75Ω
c)
d)
9 V
28.75Ω
Question: find Requiv in this circuit?
a) Requiv =R1R2/(R1 +R2)
I
I3
I2
b) Requiv =R1 +R2
c) Requiv =R1
d) Requiv =R2
0
0 = - I2 R2 +I3 R3
I2 =0, I3=I
Method II:
ε - I R1 =0
Example – The Voltage Divider
R1
R2
Vout
Representative Values
Pick α, control Vout
Example - Automobile Application
Source: http://www.smartdraw.com/examples/view/auto+wiring+diagram/
Example Question
1. Which is the Correct Circuit
Diagram, a) or b)?
a)
fuse
Current (I)
Current (I)
Total
Resistance
(RT)
12 V
Load
(R1)
Load
(R2)
Ground
Voltage
(E)
2.If the resistance of each
light bulb is 2Ω, estimate the
current I?
3.What happens if one light
bulb blows out?
b)
Summary – Kirchhoff’s Rules
(circa 1845)
• Kirchhoff’s Loop Rule
–The total change in the electric potential around any closed
circuit path must be zero
• Kirchhoff’s Junction Rule
–The current entering a circuit junction must equal the current
leaving the junction
• These circuit rules are just applications of fundamental laws of
physics
–Loop Rule – conservation of energy
–Junction Rule – conservation of charge
• The rules apply to ALL types of circuits involving ALL types of
circuit elements