Download Electric circuits, Current, and resistance (Chapter 22

Electric circuits, Current, and resistance
(Chapter 22 and 23)
Acknowledgements: Several Images and excerpts are taken from
College Physics: A strategic approach, Pearson Education Inc
Dr. Mangala Singh, 1P22/92 Brock University
Dr. Mangala Singh, 1P22/92 Brock University
Current
If electric charge (e.g.
electron) moves, we will say
an electric current, I, is set
to exist.
An electric current, I, is the
rate at which net charge (∆Q)
flows through a surface area A
∆Q
I=
∆t
Current’s units = C/S and
often written as A (Ampere)
Although current carriers (i.e., charges)
could be “+” or “-”, the direction of
current will be in the direction of a “+”
charge flow (i.e., clockwise)
Dr. Mangala Singh, 1P22/92 Brock University
I
+
A
Electrons are actual current carriers in
metals. They flow opposite to the direction
of electric field or current
Current and Current density
n = number of mobile
charges per unit volume
q = charge on each current
carriers
∆x = distance they move in
a time interval ∆t
vd = speed (called drift
velocity )
J = current density
Drift velocity vd ?
Dr. Mangala Singh, 1P22/92 Brock University
I
+
∆x = vd ∆t
∆Q = (nA∆x)q
= (nAvd ∆t )q
∆Q
I=
= nqvd A
∆t
I
J = = nqvd
A
A
Multiple choice questions
1. The charge carriers in metals are
A. electrons.
B.
positrons.
C. protons.
D. a mix of protons and electrons.
2.
3.
A battery is connected to a resistor. Increasing the resistance of the resistor
will
A. increase the current in the circuit.
B.
decrease the current in the circuit.
C. not affect the current in the circuit.
A battery is connected to a resistor. As charge flows, the chemical
energy of the battery is dissipated as
A. current.
B.
voltage.
C. charge.
D. thermal energy.
Dr. Mangala Singh, 1P22/92 Brock University
Problem: The discharge of the electric eel can transfer a charge of 2.0 mC in a
time of 2.0 ms. What current, in A, does this correspond to?
Solution:
Step 1: To solve a problem, note each and every quantities mentioned in
your problem. In this problem
Q = 2.0 x10-3 C
Time = 2.0 x10-3 s
Current = ?
Step 2: Identify relationship between these quantities
Step 3: Rearrange your equation to find out the unknown quantity
Add each and every numbers and compute.
Warning: Calculator does what you ask for….If you insert wrong numbers or in
a wrong manner, you will get a wrong answer!!! You must learn how to use your
calculator properly
I = Q/t = 2.0x10-3C/2.0x10-3s = 1A
Dr. Mangala Singh, 1P22/92 Brock University
Batteries
The potential difference between the
terminals of a battery, often called
the terminal voltage is often called
battery’s electromotive force (emf)
Wchem
∆Vbat =
=ξ
q
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Simple Circuits
The current is determined by the
potential difference and the resistance of
the wire:
∆Vwire
I=
R
battery
Dr. Mangala Singh, 1P22/92 Brock University
Ohm’s law and resistance
A current density J & the electric field E are
established in a conductor when a potential
difference (Vb-Va) is maintained across the
conductor
J = σE
------- Ohm’s Law
σ is called electrical conductivity of the materials
Ohmic materials: In these materials the ratio of J
and E is a constant “σ”. In other words, it is J
varies linearly with E
Value of conductivity of the materials are different for
different materials
Dr. Mangala Singh, 1P22/92 Brock University
V = Vb − Va
V
⇒ V = El
l
V
J =σE =σ
l
l
V=
I ⇒ V = RI
σA
l
R=
σA
E=
R=
ρL
A
Resistance & Resistivity
The resistance of a wire depends on its dimensions
(length L & Cross sectional area A) and the resistivity of
its material.
Every material has a characteristic resistivity that only
depends on the properties and composition of the
material.
Value of resistivity of the materials are different for
different materials
resistivity of the metal increase as temperature increase
resistivity of semiconductors and insulators decreases
as temperature increases
Dr. Mangala Singh, 1P22/92 Brock University
R=
ρL
A
Multiple choice
A battery is connected to a wire, and makes a current in the wire. Which of the following changes
would increase the current?
(1) Increasing the length of the wire; (2) keeping the wire the same length, but making it thicker; (3)
using a battery with a higher-rated voltage; (4) making the wire into a coil, but keeping its dimensions
the same; (5) increasing the temperature of the wire.
A.
B.
C.
D.
E.
All of the above
1 and 5
1, 4, and 5
2 and 3
None of the above
Problem 1:
A wire has resistance of 21 ohm. It is melted down and from the same volume of metal a new wire is
made, that is three times longer than the original one. What is the resistance of new wire
Problem 2:
The filament of a 100-W bulb carries a current of 0.83 A at the normal operating voltage of 120 V.
A.
B.
What is the resistance of the filament?
If the filament is made of tungsten wire of diameter 0.035 mm, how long is the filament?
Dr. Mangala Singh, 1P22/92 Brock University
Power in Circuits
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Energy and Power in Resistors
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Multiple choice
A resistor is connected to a 3.0 V battery; the power dissipated in the resistor is 1.0 W. The battery is
traded for a 6.0 V battery. The power dissipated by the resistor is now
A.
B.
C.
D.
1.0 W
2.0 W
3.0 W
4.0 W
Problem 1: An electric blanket has a wire that runs through the interior. A current causes energy to be
dissipated in the wire, warming the blanket. A new, low-voltage electric blanket is rated to be used at 18
V. It dissipates a power of 82 W. What is the resistance of the wire that runs through the blanket?
Dr. Mangala Singh, 1P22/92 Brock University
Example Problem
An electric blanket has a wire that runs through the interior. A current causes energy to be dissipated in
the wire, warming the blanket. A new, low-voltage electric blanket is rated to be used at 18 V. It
dissipates a power of 82 W. What is the resistance of the wire that runs through the blanket?
Summary
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Conservation of Current
∑ Iin = ∑ Iout
Dr. Mangala Singh, 1P22/92 Brock University
Multiple choice questions
1. Rank the bulbs in the following circuit according to their brightness, from brightest to dimmest.
A.
B.
C.
D.
A>B=C>D
A=B=C=D
A=D>B=C
B=C>A>D
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Drawing Circuit Diagrams
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Kirchhoff’s Law
∑ I in = ∑ I out
∆Vloop = ∑ Vi = 0
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Using Kirchhoff’s Laws
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Multiple choice questions
1. The diagram below shows a segment of a circuit. What is the
current in the 200Ω resistor?
A.
0.5 A
B.
1.0 A
C.
1.5 A
D.
2.0 A
E.
There is not enough information to decide.
2. The diagram below shows a circuit with two batteries and three resistors. What is the potential
difference across the 200Ω resistor?
A.
2.0 V
B.
3.0 V
C.
4.5 V
D.
7.5 V
E.
There is not enough information to decide.
Dr. Mangala Singh, 1P22/92 Brock University
Series Resistors
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Parallel Resistors
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Problems
1. There is a current of 1.0 A in the circuit below. What is the
resistance of the unknown circuit element?
2. What is the current out of the battery?
Dr. Mangala Singh, 1P22/92 Brock University
Analyzing Complex Circuits
Dr. Mangala Singh, 1P22/92 Brock University
Problems
1. In the circuit shown below:
A.
B.
Rank in order, from most to least bright, the brightness of
bulbs A–D. Explain.
Describe what, if anything, happens to the brightness of bulbs
A, B, and D if bulb C is removed from its socket. Explain.
Dr. Mangala Singh, 1P22/92 Brock University
Problems
In the circuit shown below:
A. How much power is dissipated by the 12 Ω resistor?
B. What is the value of the potential at points a, b, c, and d?
Dr. Mangala Singh, 1P22/92 Brock University
Battery and its internal resistance (r)
The current in the circuit
As the charges passes through
from – to + terminal of the
battery its potential energy
increases
I=
-Depends both external
resistance to the battery
-And its internal resistance
Due to internal resistance of the
battery, its potential decreases
by Ir, wherein I is the current.
Terminal voltage of the battery
= V+-V- = ξ-Ir
Note: Terminal voltage may exceed the emf
by an amount Ir – when the current is
opposite the emf as in the case of charging a
battery with another source of emf.
Dr. Mangala Singh, 1P22/92 Brock University
Power dissipated, P, in the load
resistance R
2
P=I R=
ξ2
(R + r)
2
R
ξ
R+r
Problems:
1. Show that maximum power lost in the load resistor R occurs
when R =r, i.e., when the value of load resistance matches the
internal resistance of the battery.
Dr. Mangala Singh, 1P22/92 Brock University