WileyPLUS Assignment 1
Chapters 18 and 19
Due: Wednesday, January 28 at 11 am
Notice on PHYS1030 home page with WileyPLUS link
Q18.52: Gauss’ Law - will not be on midterm or final
Week of Jan 26 – Jan 30
Experiment 2: Wheatstone Bridge
Monday, January 26, 2009
52
Kirchhoff’s Rules
1) Junction Rule:
! The sum of currents entering a junction is equal to the sum of
currents leaving it (conservation of charge and current).
2) Loop Rule:
! Around any closed loop, the sum of potential changes is equal
to zero (conservation of energy).
The potential decreases when you go with the flow of current
through a resistor.
– use the “conventional” flow of current
Monday, January 26, 2009
53
Experiment 2: Wheatstone Bridge
To verify Stefan’s Law: power radiated (light, heat) by a heated object
is proportional to T4. T = temperature in Kelvin.
Heated object is the filament of a lamp heated by an electric current.
In a steady state, power radiated = electrical power supplied.
Power supplied is P = V2/R.
Resistance R of the heated filament is measured with a Wheatstone
bridge. V is measured with a voltmeter.
The temperature of the filament is determined from its resistance.
R = R0 [1 + !(T - T0)]
Is P proportional to T4 ?
Monday, January 26, 2009
54
Experiment 2: Wheatstone Bridge
ce
an
st
e
l
ab
i
r
Va
si
e
r
Ig = 0
C
Voltmeter
A
Galvanometer. Reads zero when:
D
R1 R4
R2
V2
Then, P =
Rx
V
B
Rx =
(then
VA = VB)
From Rx, find temperature of
the filament from graph
provided in lab manual.
Is P proportional to T4 ?
Monday, January 26, 2009
55
Experiment 2: Wheatstone Bridge
As Ig = 0 when bridge balanced,
I1 continues from C to A to D
I2 continues from C to B to D
A
r
I1
ia
r
Va
e
bl
e
nc
a
t
is
I1
s
e
CABC: -I1R1 + I2Rx = 0
Ig = 0
C
D
I2
(1)
ADBA: -I1R2 + I2R4 = 0 (2)
From (1) and (2):
I2
I1
Rx
R4
=
=
I2
R1
R2
B
So, Rx =
R1 R 4
R2
Monday, January 26, 2009
56
20.79: Wheatstone Bridge
Find potential difference
between B and D
I1
I1 – I2
I1
What is the equivalent
resistance between A and C?
I2
I – I1
I1 – I2
I
I
I – I2
I – I1
Req = V/I = 20/I
I – I2
I
What is I ?
F
Monday, January 26, 2009
I2
E
57
Clicker Question
The circuit in the drawing contains three identical resistors.
Consider the equivalent resistance between the pairs of points
a and b, b and c, and a and c.
Rank the equivalent resistances in decreasing order.
A) a to b;
B) a to c;
C) a to c;
D) b to c;
E) a to b;
b to c;
b to c;
a to b;
a to c;
a to c;
a to c
a to b
b to c
a to b
b to c
Answer C)
Monday, January 26, 2009
58
20.104: Find the current through the 2 ! resistor and the voltage
V of the battery to the left of it.
A
3A
F
Monday, January 26, 2009
B
3A
3–I
C
I
I
E
D
59
A galvanometer
Full scale deflection (fsd) = 0.1 mA
(for example)
Meter
Rc
Resistance of galvanometer coil
A torque is exerted on the coil when a current flows through it. The
rotation of the coil is resisted by a spring. The angle of rotation is
proportional to the current in the coil.
Monday, January 26, 2009
60
Mode of operation of an ammeter to measure current
The ammeter is inserted into the
circuit to measure the current
flowing around the circuit.
R
Monday, January 26, 2009
Ideally, the ammeter should have
negligible resistance so as not to
affect the current being
measured.
Rc << R
61
Measurement of current –!switching scales
Ammeter – a bypass (shunt) resistor is placed
in parallel with a galvanometer to limit the
current through the galvanometer to no more
than that for full scale deflection (fsd)
To measure 60 mA when the fsd is 0.1 mA:
VAB = 0.1RC = 59.9R (mV)
galvanometer
(fsd)
So, for the shunt resistor:
R = RC
0.1
= 0.00167RC
59.9
To measure current I: R = RC
fsd
I − fsd
Monday, January 26, 2009
62
Prob. 20.107/83: A galvanometer has a coil resistance of 12 ! and a
full scale deflection current of 0.15 mA. It is used with a shunt
resistor to make an ammeter that registers 4 mA at full scale
deflection.
Find the equivalent resistance of the ammeter.
• Work out what is the potential difference across the
galvanometer when 0.15 mA flows through it.
• Find what the shunt resistance needs to be to produce an equal
potential difference when the current flowing through the shunt
resistor is 4 – 0.15 mA.
Monday, January 26, 2009
63
Mode of operation of a voltmeter to measure
voltage
The voltmeter is attached to two points
of a circuit between which the potential
difference is to be measured.
Ideally, the voltmeter should have very
large resistance so as to draw very little
current from the circuit being studied.
Monday, January 26, 2009
64
Measurement of voltage
V = 100 V
Coil resistance
R
galvanometer
I f sd = 0.1 mA
RC = 50 !
I = 0.1 mA
A resistor is put in series with the
galvanometer to limit the current to
that giving full scale deflection (fsd)
at the desired voltage.
To measure 100 V when fsd is at 0.1 mA and the coil resistance RC = 50 !:
V = I f sd (R + RC )
That is, R =
(R = resistance put in series with the galvanometer
to convert it to a voltmeter)
V
100 V
− RC =
− (50 !) = 999, 950 !
I f sd
0.1 × 10−3 A
R ! 1 M!
Monday, January 26, 2009
65
Prob. 20.82: Voltmeter A has an equivalent resistance of 2.4 " 105 !
and a full scale voltage of 50 V.
Voltmeter B uses the same galvanometer. It has an equivalent
resistance of 1.44 " 105 !. What is its full scale voltage?
Monday, January 26, 2009
66
Ammeters and Voltmeters
Ammeter: put a resistance in
parallel with the galvanometer
to divert current from the
meter
Rc
Voltmeter: put a resistance in
series with the galvanometer
to limit the current through
the meter
Ifsd
R
R
I
I - Ifsd
Ifsd Rc = (I - Ifsd) R
Monday, January 26, 2009
I
Ifsd
RC
V
V = Ifsd (R + Rc)
67