Course and Section_______ Names ___________________________
Date___________ _________________________________
This lab deals with circuits involving resistors, capacitors and inductors in which the currents and voltages vary sinusoidally in time.
Equipment
1 function generator (PC Scope software)
1 digital multimeter and leads
1 decade resistance box
1 capacitor (nominally 0.1
μ
F)
1 inductor (nominally 10 mH)
1 mini-jack to banana plug (black, red, blue) cable, 2 alligator clips
Introduction
If the current through a passive component is given by i ( t )
=
I sin(
ω t )
=
I sin( 2
π f t )
. (1) then the voltage across the component also varies sinusoidally, but with a phase that depends on the component. For a resistor the voltage is in phase with the current. For a capacitor the voltage lags the current by 90 o
, and for an inductor the voltage leads the current by 90 o
. The peak (or rms ) current-voltage relationships for a resistor, capacitor and inductor are
V
R
V
C
=
IR
=
IX
C
=
I /(
ω
C )
(2)
(3)
V
L
=
IX
L
=
I
ω
L
(4)
Series RC Circuit
In a series RC circuit, since the currents are the same then the voltages across R and C are 90 o
out of phase.
Consequently, the total voltage across the combination is
R
C
V
=
I R
2 +
X
C
2
(5) v(t) = V sin(
ω
t+ ϕ
)
Figure 1

Series RL Circuit
In a series RL circuit, the voltages across R and L will also be 90 o
out of phase. Thus,
V
=
I R
2 +
X
L
2
Series RLC Circuit
(6)
R v(t) = V sin(
ω
t+ ϕ
)
Figure 2
L
In a series RLC circuit, since the voltage across L leads the current by
90 o
and the voltage across C lags the current by 90 o
, then the voltages across L and C are 180 o
out of phase. Consequently, we have
V
=
I R
2 +
(
X
L
−
X
C
)
2
. (7)
We can write this as V
=
I Z , where Z
=
R
2 +
(
X
L
−
X
C
)
2
is the
R C L v(t) = V sin(
ω
t+ ϕ
) circuit impedance. In the series RLC circuit, the current will be a maximum when the impedance is a minimum, that is, when X
L
= X
C
, or f
= f
0
=
2
π
1
LC
(8)
Figure 3
1.
In a series RC circuit, V
R
and V
C
are measured as a function of frequency. Do you expect V
R
and V
C
to increase, decrease, or remain constant as you change f ? Show your predictions by making a sketch of V
R and V
C
versus f .
Solution:
Using equation (5) above we have for a series RC-circuit the following equation for the amplitude of the current:
V
I
=
(5’)
R
2 +
X
C
2
We can use this to calculate the voltage across the resistor (which is in phase with the current trough the resistor):
V
R
=
IR
=
R
2
VR
+
X
C
2
with X
C
=
ω
1
C
=
2
π
1 f C
for small frequencies f the voltage drop will be mainly across the capacitor and thus V
R
will go to zero as f decreases. For very large frequencies however will approach zero and thus the voltage drop will be mainly across the resistor and
=
V . f lim
→ ∞
V
R
X
C
To get the voltage drop across the capacitor we use again equation (5’) for the current through the capacitor and get for the voltage drop across the capacitor:
V
C
=
IX
C
=
VX
C
R
2 +
X
C
2

Below is a graph that shows the expected frequency dependence of V and
R
V
C
Series   RC ‐ circuit
1.20E+00
1.00E+00
8.00E
‐ 01
6.00E
‐ 01
4.00E
‐ 01
2.00E
‐ 01
V_C
V_R sqrt(V_R^2+V_C^2)
0.00E+00
0 2000 4000 6000 8000 f   [Hz]
2.
In a series RL circuit the rms voltage across R is 30 V and the rms voltage across L is 40 V. What is the rms value of the voltage across the RL combination?
Solution:
Starting with equation (6)
V
=
I R
2 +
X
L
2 =
( ) ( )
L
2 =
V
R
2 +
V
L
2
thus we get for the rms value across the RL combination:
V
=
30
2 +
40
2 =
50 V
In this experiment, you will use the function generator feature of the PC Soundcard Oszilloscope program for your voltage source and the handheld multimeter to measure the voltages. (You will not use the PC oscilloscope.) The Oszilloscope program can be accessed from the ‘Scope’ icon on the desktop. Use the red and black leads of the cable for the function generator output and make sure the volume of the PC soundcard is not set on mute.
Series RC Circuit
Wire the series RC circuit as shown in Fig. 1. Use the decade resistance box and set R = 500
Ω
. (Note:
Measure the resistance to ensure that R is approximately the set value. If it is way off, then this particular setting is damaged. Then pick another setting (e.g., 400
Ω
or 600
Ω
).) Turn Channel 1 of the ‘scope’ function generator on, set it to ‘sine’, and set the amplitude to ‘1’. (Note: The actual function generator output voltage will depend on the load.) Now use the multimeter to measure V
R
, V
C and V (function generator voltage) as a function of frequency over the range from 100 Hz to 8000 Hz. Use steps of 100,
500, 1000, 1500, 2000, … , 8000 Hz.
Enter your data into Excel. In a separate column calculate V
R
2 +
V
C
2
. Now on a single graph, plot V
R
,
V
C
, V, and V
R
2 +
V
C
2 as a function of f .
Analysis:
1.
Are your measurements in qualitative agreement with your answer to Preliminary Question 1? Explain.

The results should look something like this – but with fewer data points:
Series   RC ‐ circuit
1.20E+00
1.00E+00
8.00E
‐ 01
6.00E
‐ 01
4.00E
‐ 01
2.00E
‐ 01
V_C
V_R sqrt(V_R^2+V_C^2)
0.00E+00
0 2000 4000 6000 8000 f   [Hz]
2.
How does your source voltage, V , compare with V
R
2 +
V
C
2
? Why don’t the voltages add as V = V
R
+
V
C
?
Solution:
Just adding the voltages (like in a DC circuit) does not take into account the phase shift. As there is a 90 o phase shift one gets equation (5):
V
=
I R
2 +
X
C
2 =
( ) (
IX
C
)
2 =
V
R
2 +
V
C
2
3.
Note the frequency at which V
R
= V
C
. Use this value of f and your known value of R to calculate C .
f = 3.2 kHz C = 0.1
μ
F
Solution:
V
R
=
IR
=
V
C
=
IX
C
as the current I is the same this happens for
R
=
X
C
=
2
π
1 f C
solving for C gives:
C
=
2
π
1 f R
=
0.1
μ
F
Series RLC Circuit
Now add the inductor in series with the above circuit to form a series RLC circuit. Change the resistance setting from 500
Ω
to 50
Ω
. (Check the value of the resistance with the multimeter, if 50
Ω
is not working try 40
Ω
or 60
Ω
). Measure the voltage across the resistor as a function of frequency from 100 to 8000 Hz.
Use steps of 100, 500, 1000, …, 8000 Hz. Note the approximate frequency where V
R
reaches its maximum value. Now add a few additional measurements at frequencies near this value to better define your resonance.

Analysis:
1.
Make a graph of V
R
versus f and determine the resonance frequency, f
0
. Use Eq. (8) and your previously determined values of R and C to calculate L .
Solution:
Starting from equation (8) f
0
=
2
π
1
LC
And solving for L gives:
L
=
1
C
(
2
π
1 f
0
)
2 from the graph one gets f
0
=5kHz and thus L=10mH
Series   RLC ‐ circuit
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
V_R
V_R/sqrt(2)
0.2
0.1
0
0 1000 2000 3000 4000 5000 6000 7000 8000 f   [Hz]
2.
Set your frequency to the resonance value. Now measure V
L
, V
C
, and V
LC
, the voltage across the LC combination. Ideally, at resonance the voltages across L and C are 180
0
out of phase and exactly cancel so that V
LC
= 0. Presumably, this is not what you observe. The reason is because the inductor has a resistance R
L
that is effectively in series with L . So, at resonance you would expect that V
LC
= IR
L
. Use the multimeter (resistance setting) to measure R
L
.
Measure the source voltage V and calculate the current using I = V/(R+R
L
).
Then calculate IR
L
and compare with your measured value of V
LC
.
Solution:
For this particular coil R
L
= 5 Ohm.
Thus for the current at resonance one expects: I = V/(R+R
L
)=18 mA and thus V
LC
= IR
L
= 0.1 V, see graph below:
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
V_LC
2000 3000 4000 5000 f   [Hz]
6000 7000 8000
The frequency dependence of V
L
, V
C
and V
LC
should look like this

4
3
2
1
6
5
V_LC
V_L
V_C
0
3000 4000 5000 f   [Hz]
6000 7000 8000
One interesting observation is that although the circuit is only driven with a 1 volt source the maximum voltage across the capacitor and inductor both exceed this value by almost a factor of 6. However, they are out of phase with each other, which results in a much smaller voltage drop across the combination of the two (V
LC
). For a quantitative estimate of the maximum voltage drop across the capacitor and inductor see part 3 below.
3.
The ‘quality factor’, Q, of the resonant circuit is defined as
Q
= f
0
Δ f
, (9) where
Δ f is the width of the resonance curve measured between points where the power dissipation is ½ the value at resonance. Since P = V
R
2
/R, the half-power points correspond to the points where
V
R
=
V
R
(max) / 2 .
Series   RLC ‐ circuit
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
3500 4000 4500
V_R
V_R/sqrt(2)
5000 f   [Hz]
5500 6000 6500
Power
18
16
14
12
10
8
6
4
2
0
3500 4000 4500 5000 f   [Hz]
5500 6000 6500

From these graphs one has:
Δ f
≈
850 Hz and thus Q=5.9
It can be shown that Q can be calculated using the expression
Q
=
2
π
R f
0
L
(10)
Measure Q using Eq. (9) and compare with your calculated Q using Eq. (10). In Eq. (10) R should be the total resistance of the circuit (decade box plus coil resistance).
Using the values for f
0
, L and R obtained earlier one has:
Q=5.8 close to the value extracted from the above graphs. One notes that for the maximum voltage across the capacitor one has: ,
,
and similarly for the maximum voltage across the inductor:
in good agreement with the graphs in 2.