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PH-1120 Lab Notes
Experiment No. 2
Direct Current Measurements
by R. Long
PH-1120: General Physics – Electromagnetism
Laboratory Procedure.
Part I. Calculation of resistance using the direct application of Ohm’s Law.
1) Set Rload = 0 (decade box). The resulting circuit can then be redrawn more simply as
in the diagram below.
+
mA
Adjustable
Voltage
Source
V
Ru
Sw
2) Set V to 1 volt by adjusting the rheostat while holding down the switch key (Sw)
briefly.
3) Release Sw to let Ru to cool briefly.
4) Press Sw briefly and read ImA, the current through the unknown Ru.
5) Record V and I (Table I).
6) Return to step 2 and increase V by 1 volt. (Note: Vmax = 10 volts).
Table I. Current vs. Voltage
V (volts)
I (mA)
Part II. Power transfer technique for determining Ru.
This part of the experiment makes use of the fact that the battery will deliver
maximum power to the load resistor when the value of the load resistor equals the internal
resistance of the battery. A plot of power dissipated in the load resistor (R load) as a function
of the value of Rload will yield a curve with its maximum occurring when R load equals the
internal resistance of the battery. This is not a practical method for determining the actual
internal resistance of a battery or the power supply because such internal resistances are
normally very small (less than one Ohm).
To employ this technique we shall simulate a battery with a large internal resistance
by placing the unknown resistor of Part I in series with the power supply and consider it as
the internal resistance of the battery.
1) Set Rload = 80 Ohms (decade box).
2) Set the rheostat for Vmax = 10 volts (regulated). (Keep the voltage at 10 volts
throughout the experiment by readjusting the rheostat as necessary.) Now, when the
switch key (Sw) is pressed, points A and B act like the terminals of a battery with a
10 volt emf and an internal resistance Ru. The resulting circuit can be redrawn as in
the diagram below.
I
A
Є
mA
Rload
Sw
Simulated
Battery
B
External Circuit
connected to
battery
terminals
A delivered
and B to the load resistor. Record only
3) Press Sw briefly and read ImA, the current
Ru
Rload and ImA and the uncertainty in your reading for ImA in Table II.
Table II. Power vs. Load Resistance
Rload
ImA
Δ ImA*
Δ V **
P
Δ Rload +
Eq. (1)
ΔP
Eq. (2)
*
You cannot read the milli-ammeter exactly so you must estimate the inaccuracy or
uncertainty, Δ ImA, in your reading.
Example: I = 20.0 A ± 0.1 A. Note that these values are illustrative only and are not
related to this experiment.
**
Δ V is the uncertainty in reading the Voltmeter.
+
The decade box resistance is accurate to ± 0.1%. Use this information to estimate the
uncertainty, Δ Rload, in Rload.
Results:
Part 1: Determine the value of Ru graphically from a plot of I as a function of V. The
independent variable, I, is to be plotted as the ordinate and V, the independent variable as
the abscissa. The unknown resistance, Ru, is obtained as the reciprocal of the slope of the
best straight line fit through the datum points.
Part IIa: The power dissipated in Rload can be calculated using the following relationship:
P = I2 Rload
(1)
Plot a curve of P as the ordinate versus Rload as the abscissa. However, before
attempting to draw the curve to find the maximum you are asked to determine the
uncertainty in the values of P so that you can draw error bars around the datum points,
indicating this uncertainty.
To determine the uncertainty ΔP in P we express the uncertainty, in P, in terms of
the uncertainties in I and Rload = Rl.
P + ΔP = (I +ΔI)2(Rl + ΔRl) = [I2 + 2IΔI + (ΔI)2] (Rl + ΔRl),
P + ΔP = I2 Rl + I2ΔRl + 2IΔI· Rl + 2IΔI· ΔRl + (ΔI)2 Rl + (ΔI)2 ΔRl.
Note: In these expressions the uncertainties ΔI and ΔRl may be either positive or negative.
Since P = I2Rl,
ΔP = I2ΔRl + 2IΔI· Rl + 2IΔI· ΔRl + (ΔI)2 Rl + (ΔI)2 ΔRl.
Because ΔI and ΔRl are small compared to I and Rl, we can neglect any products
containing two or more of these delta quantities. Then,
ΔP = I2ΔRl + 2IΔI· Rl.
Divide by P = I2Rl to get
(ΔP/P) = (ΔRl)/ Rl + 2(ΔI)/ I.
The quantities ΔP/P, ΔI/I and ΔRl/Rl are called the fractional uncertainties of P, I and Rl,
respectively. The fractional uncertainty in P is equal to the fractional uncertainty of R plus
twice the fractional uncertainty in I. (The fractional uncertainty in I is doubled because I is
squared in the expression for power.)
We can estimate the magnitudes of the uncertainties of ΔRl and ΔI in any given
measurement but we do not know their signs. To avoid placing unwarranted trust in our
values of P, we want to know the possible extreme values of ΔP/P. The two worst cases
are
(ΔP/P) = ± (|ΔRl|/ Rl + 2|ΔI|/ I).
(2)
The upper sign corresponds to both ΔRl and ΔI positive and the lower sign to both
negative. The worst uncertainty in P is then
ΔP = ± (|ΔRl|/ Rl + 2|ΔI|/ I) · P.
(3)
On our graph of P vs. Rl we show the uncertainty in each value for P by means of an
error bar extending above and below each P datum point by an amount ΔP, as depicted in
the following sketch.
Power
P + ΔP
Error Bar
P=I2RL
P - ΔP
Data Point
RL
Load Resistance
Draw a smooth curve that best fits the P vs. Rload datum points. ( It should be as near
as possible to the datum points and with in the error bars, but should be smooth and not
drawn by connecting points.) The value of Ru is obtained as the value of Rload at the point
of maximum P.
Part IIb: Another way to determine Ru from the power data is by algebraic manipulation of
the power equation. At maximum power the dissipation in Rload, Rload = Ru and I = V/( Ru +
Rload) = V/(2 Ru). Substituting this into the power equation, we find that
Pmax = (V/(2 Ru))2 Ru, or
Ru = V2/(4Pmax) .
(4)
Here Ru is obtained in terms of the maximum power determined in part IIb.
Analysis:
Part I: Estimate the fractional uncertainty, ΔRu/Ru for this method. The differential of
Ohm’s Law,
Ru = V/I,
(5)
dRu = dV/I – V (dI/I2).
(6)
is
Dividing eq, (6) by eq. (5) yields
dRu/ Ru = dV/V - dI/I.
(7)
The worst case fractional uncertainty is obtained when both terms on the right side have
the same sign;
dRu/ Ru) worst = ± ( |dV/V| + |dI/I| ).
(8)
Since the straight line was supposed to average all the experimental points, it seems
consistent with the spirit of this average to evaluate eq. (8) with V and I from the middle of
their respective ranges; ΔV and ΔI are the same as in Table II.
Part IIa: Estimate the un certainty ΔRu for this method. There are other possible averaging
curves for P vs. Rload than the one you have used. What is the greatest amount by which
you can shift the location of Pmax along the Rload axis from your chosen location by drawing
a different smooth averaging curve within the error bars? This is your estimate of ΔRu for
part IIa.
Part IIb: Estimate the fractional uncertainty ΔRu/Ru for this method. The differential of eq.
(4) is
dRu = (VdV) / 2Pmax - (V2dPmax) / 4(Pmax)2.
(9)
Dividing eq. (9) by eq. (4) yields
dRu/Ru = 2 (dV/V) - dPmax/Pmax .
(10)
The worst case fractional uncertainty is obtained when both terms on the right side have
the same sign;
Δ Ru/Ru )worst = ± ( 2|ΔV/V| + |ΔPmax/Pmax| ) .
(11)
Evaluate this fractional uncertainty using ΔPmax and ΔV from Table II near to the value of
ΔPmax.
Conclusions.
Table III Comparison of the results.
Method
I
IIa
IIb
Ru Ohms
Δ Ru/Ru
Δ Ru Ohms
The Grade for your Laboratory Report will be based on the following:
1) Cover Page attached (Stapled) and complete (see instructions for Lab #1).
2) Tables I, II and III, including a sample calculation of each derived (non-measured)
quantity. Be sure to include proper units.
3) Graphs for part I and IIa.
4) Conclusions based on table III. Which technique gives the most precise result? The
most uncertain result? What value would you assign to Ru ? Explain?
5) Original of your data sheet.
First revised
JFW B’87
Second Revision RG B’90
Third Revision
B’93
Most recent
D’05