Download Ch. 20

Ch. 25
Electric Current and DC
Circuits
Chapter Overview
Definition of Current
 Ohm’s Law
 Resistance – Conduction in Metals
 Kirchhoff’s Laws
 Analysis of DC Circuits
 RC Circuits

Current
Up to this point we have been concerned
with charges that don’t move – Static
 When charges do move, then an electric
current flows
 Current, usually denoted by the letter i is
that rate at which charge moves. In other
words how much charges flows past a
point per time

Current






i = current
q = charge
t = time
SI Units – ampere
Symbol A
Fundamental Unit
(More on this later)
q
i
t
The ampere is a fundamental unit,
so the coulomb is a derived unit.
Express the coulomb in terms of
fundamental units
1.
2.
3.
4.
1
2
A/s
A·s
s/A
None of he above
3
4
5
0%
1
0%
0%
2
3
0%
4
Batteries are often rated in
amp·hours. What type of quantity
does an amp∙hour represent?
Current
charge
Electric Potential
Capacitance
None of the above
1.
2.
3.
4.
5.
1
2
3
4
5
0%
0%
0%
0%
1
2
3
4
0%
5
Ex. How many coulombs of
charge are stored in 60 A·hr
battery?
Solution
i=Δq/Δt
 Δq=iΔt
 Δq=60 A x 1 hr
 =60A x 1 hr x 3600 s/1 hr =216000 C

Ex. 10000 protons fIow through
a detector every .05 s. What is
the current flowing through the
detector?
Soln.
Protons



Detector
Current i=Δq/Δt
i=10000x1.602x10-19 C/.05 s
i= 3.2 x 10-14 A
Microscopic View of Current

At the microscopic level, current is made by
individual charges moving at a speed vd in the
material
x
vd
q

A charge will travel the distance x in a time given
by t = x/vd
Microscopic view of current
The total amount of charge that flows
through the gray shaded volume in time Δt
is ΔQ = nqV where n is the number of
charges per volume, V is the volume of the
gray shaded area, and q is the charge of
an individual charge
 V = xA = vdΔtA
 So ΔQ = nqvdAΔt

Drift Velocity

I = ΔQ/Δt = nqAvd

vd is the drift velocity. It represents the average
speed of charges in conductor

Ex. A copper wire has a radius of .50 mm. It
carries a current of .25 A. What is the drift
velocity of the electrons in the wire. Assume 1
free electron per atom. (ρcu = 8.93 g /cm3)
There will be Avagadro’s
number of charges in 1 mole of
copper. the molar mass of
copper is 63.5 g. The volume
of 1 mole of copper is
3
V = 63.5 g/8.93 g/cm
= 7.31 cm3 = 7.31 x 10-6 m3
n = NA / Vmol
= 6.022 x 1023/7.31 x 10-6 m3
28
3
= 8.24 x 10 electrons /m
vd = i/nqA
= .25 A/(8.24 x 1028/m3 x
1.602 x 10-19 As x 3.14 x(5 x 104 m)2) = 2.4 x 10-5 m/s
Ohm’s Law

What factors determine how much current
will flow in a circuit (BRST)
Ohm’s Law

Potential Difference and the material
properties determine current that
flows in a circuit
V
i
R
Ohm’s Law




Current is proportional to potential difference
“Resistance” limits the amount of current that
flows
Experimental Relationship found by Georg
Simon Ohm
Not always true – e.g. diodes, transistors,…
Resistance
Units – ohm denoted by Ω
 What is the ohm in terms of V and A
 What is the ohm in terms of fundamental
units

V
R
i
Units

Ω=V/A

Ω = kg m2 /(A2 s3 )
Resistivity

What factors determine the resistance of a
piece of metal? (BRST)
Resistivity
Length – L
 Cross sectional Area – A
 resistivity ρ – material property (see table
in text p. 792)

L
R
A
Two different metals with different
resistivities have the same length. Which
metal will have the higher resistance?
The metal with the greater ρ
The metal with the smaller ρ
The resistances will be the
same
Cannot be determined
1.
2.
3.
4.
0%
1
2
3
4
5
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0%
0%
2
3
0%
4
Temperature Dependence of
Resistance

If you measure the resistance of a light
bulb cold and then measure it when it is
glowing, do you get the same resistance?
(BRST)
Temperature Dependence of
Resistance
Resistance increases with increasing
temperature for metals
 It decreases with increasing temperature for
semiconductors
 For conductors
R = R20(1 + α(T- 20 C°))
α is the Temperature coefficient of resistance (see
p. 792)

Ex. Find the temperature of the
filament of a light bulb (assume
W) by measuring the resistance
when cold and glowing.
Power

When we apply a potential difference
across a resistor, it gets hot

What determines the power given off by a
resistor?
Power

Work = qΔV

Power = Work/time

P = qΔV/t but q/t = i

P = iΔV
Power

Combine the expression for electric power
P = iΔV with Ohm’s law ΔV = iR

P = iΔV = i2R = (ΔV )2/R
Ex. A 100 W bulb is designed to emit
100 W when connected to a 120 V
circuit. a) Draw a sketch and a
schematic. b) What is the resistance of
the bulb and the current drawn when
connected to a 120 V outlet?
c) Assuming the resistance doesn’t
change, what would be the power output
if the bulb was connected to a 240 V
circuit?
The electric bill

Power is a rate – it tells you how much
energy per time is being used
The electric company bills you in units
of kw∙hr. What is a kw∙hr? (TPS)
It is a unit of power
It is a unit of energy
It is a unit of current
It is a unit of
potential difference
Not enough
information given
1.
2.
3.
4.
5.
1
2
3
4
5
0%
1
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3
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0%
5
The electric company bills you in
units of kw∙hr. What is a kw∙hr?
(TPS)
It is a unit of power
It is a unit of energy
It is a unit of current
It is a unit of
potential difference
Not enough
information given
1.
2.
3.
4.
5.
1
2
3
4
5
0%
0%
0%
0%
1
2
3
4
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EX. How many joules are in a
kW∙hr?
For the circuit shown below how
does the current flowing through A
compare to that flowing through B
It is the same
It is greater at A
It is greater at B
It cannot be
determined
1.
2.
3.
4.
3.0 Ω
V
6.0
Ω
0%
1
2
3
4
5
B
A
1
0%
0%
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4
You connect two identical resistors in series
across a 6.0 V battery. How does the current
in the circuit compare to that when a single
resistor is connected across the battery
1.
2.
3.
4.
There is no difference
The current is twice as large
The current is ½ as large
Cannot be determined
0%
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2
3
4
5
1
0%
0%
2
3
0%
4
Resistors in Series
When we add resistors is series, the
current decreases since the resistance
increases
 We define an equivalent resistance as a
single resistor which produces the same
current when attached to the same
potential as the combination of resistors

Series equivalent
Equivalent Series Resistance
We want i to be the same
 V= iReq
 V1 = iR1, V2 = iR2, V3 = iR3
 V = V1 + V2 + V3 = iR1 + iR2 + iR3
 iR1 + iR2 + iR3 = i(R1 + R2 + R3) = iReq
 So Req = R1 + R2 + R3

Equivalent Series Resistance

How would this result change if there were
four resistors in series?

In general as more resistors are added in
series, the resistance increases so the
current decreases
Ex. a) Find the equivalent resistance for the following
circuit. b) Find the current In the circuit c) Find the potential
drop across each resistor d) Find the power dissipated by
each resistor. e) Find the power Supplied by the power
supply.
You connect two identical resistors in parallel
across a 6.0 V battery. How does the current
supplied by the battery compare to that when a
single resistor is connected across the battery?
1.
2.
3.
4.
There is no difference
The current is twice as
large
The current is ½ as large
Cannot be determined
0%
1
2
3
4
5
1
0%
0%
2
3
0%
4
Resistors in Parallel
When we add resistors is parallel, the
current increases
 The effective resistance must then
decrease
 How can that be? (BRST)

Resistors in Parallel

There are more branches for current to
follow in a parallel circuit, so current can
be larger
Resistors in Parallel

What is the same for the three resistors
shown? (GR)
Resistors in Parallel
The potential difference across each
resistor is the same
 Define i1 = V/R1, i2 = V/R2, i3 = V/R3


How do the currents combine?
Parallel Equivalent Resistance

We define the equivalent resistance as a
single resistor that will draw the same
current from the power supply
Parallel Equivalent resistance


The currents add. Why?
i = i1 + i2 + i3 = V/R1 + V/R2 +
V/R3 = V/Req
V
V V V



R eq R 1 R 2 R 3
1
1
1
1



R eq R 1 R 2 R 3
a) Find the equivalent resistance.
b) Find the current flowing through
each resistor. c) Find the current
supplied by the power supply
Find the equivalent resistance for
the following network
Kirchhoff’s laws
The rules for series and parallel
resistances are examples of Kirchhoff’s
Laws
 Voltage Law - Sum of the potential
differences around a closed loop is 0
 Current Law- the sum of currents at a
node is 0

Voltage Law

A loop is a closed path in a circuit
Current law

A node is a point in a circuit where several
wires join
For the circuit shown below a) choose currents for each
branch of the circuit. b) For the choice of currents you’ve
made, label the higher potential side of each resistor with a
+ and the lower potent side with a – c) Use Kirchoff ‘s laws
to write a closed system of equations for the currents. d)
Solve for the currents. e) Find the potential difference
across each resistor. f) Find the power dissipated by each
resistor. g) Find the power supplied by each’ power supply
RC Circuits

A resistor in series with a capacitor makes an RC circuit

RC circuits have many applications – e.g. camera
flashes
RC Circuits - Charging

Use Kirchhoff’s Voltage Law to analyze the
circuit shown below
V
R
C
RC Circuits
V – VR – VC = 0
 V – iR – Q/C = 0


But i = dQ/dt

V - R dQ/dt – 1/C Q = 0
RC Circuits

Kirchhoff’s Voltage Law gives a differential
equation for the charge

dQ/dt = V/R - Q/RC
Assuming the capacitor is initially discharged the
solution (You’ll work it out in lab)
Q = CV(1 – e-t/(RC))


Charging a Capacitor
Vc (V)
Charging a Capacitor with V = 6 V andRC =
5s
7
6
5
4
3
2
1
0
When t = RC = 5 s,
Vc = .63 * 6V =3.78 V
0
5
10
15
Time (s)
20
25
Current in a Charging RC Circuit

i = dq/dt = V/R e-t/RC

The current exponentially decays with the
same time constant
Discharging a Capacitor
Assume that the capacitor is initially
charged with Q0 = CV0
 What will happen when the switch is
closed

R
C
Discharging a Capacitor
-R dq/dt – q/C = 0
 dq/dt = -q/(RC)


A solution is q = CV0 e-t/(RC)
Solution is exponential decay
 RC is the time constant

Ex. a) What fraction of the
original charge remains when
t = RC? b) At what time is
charge reduced to a fraction f of
the initial amount?
Current in a Discharging RC
Circuit

i = dq/dt = -V/R e-t/RC
The current exponentially decays with the
same time constant
 It flows in the opposite direction
