# Albany Medical College The RLC Circuit Lab Report Experiment 11: Physics Answers 2021

## Albany Medical College The RLC Circuit Lab Report Experiment 11: Physics Answers 2021

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Albany Medical College The RLC Circuit Lab Report Experiment 11

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Florida Institute of Technology
PHY 2092 IN – LAB WORKSHEET
Experiment 11 The RLC Circuit
Name: ________________________
Date: __________
Section # ________
Read the following and fill in the blank lines with the correct equation or response. This
worksheet will be graded out of 11 points and averaged in with all other worksheets.
First, we apply Kirchhoff’s voltage rule to a series RLC circuit driven by an oscillating voltage
given byV0 cos (wt). Kirchhoff’s rules states that the voltage drops (losses of energy) across the
resistor, inductor and capacitor equal the applied voltage (the energy source). In other words,
VL + VR + VC + V0 cos (wt) = 0
(1)
Faraday’s Law gives VL = – L (dI / dt)
Ohm’s Law gives
VR = ________.
The definition of capacitance gives VC = ________.
All three terms have minus signs to indicate the current loses voltage (energy) when it passes
through these elements. Also, the cosine driving voltage can be written as a complex
exponential using Euler’s relation:
eiwt = cos(wt) + i sin(wt)
where i = (-1)1/2
Since only the real part of complex exponentials are physically meaningful, we can legitimately
write V0 coswt = V0 eiwt. Substituting all this into Eqn. (1) yields the following.
L
d 2q
dq 1
+ q = V0 eiω t
2 + R
dt
dt C
(2)
Next, take the time derivative of Eqn. (2) and write the result below. The first term is given.
d3q
L 3 +
dt
____________________________
(3)
Now, Rewrite Eqn. (3) by using dq / dt = I. This returns the equation to one involving only
second-order time derivatives. Whew!
________________________________________________________
11 – 4
(4)
Florida Institute of Technology
Since the applied voltage oscillates, we’ll guess that another sinusoidal function would solve this
differential equation. So, consider a trial solution of the form I(t) = Io eiwt where Io may be a
complex number. To see if this guess is a correct, we substitute this into Eqn. (4) which gives
1 ⎤ iωt

2
iωt
= iωV0 e
⎢⎣− Lω + iRω + C ⎥⎦ I0 e
(5)
By dividing throughout by iw eiwt and using 1 / i = -i, we obtain a form of Ohm’s Law found in
Eqn. (5). Use the next two lines to perform this division and rearranging.
[
____________________________ I0 = V0
]
(6)
[
____________________________ I0 = V0
]
(7)
Then you should arrive at Eqn. (8):

⎛ ωL − 1 ⎞⎤ I = V
R
+
i

0
⎢⎣

ωC ⎠⎥⎦ 0
(8)
So, the trial solution is correct if Eqn. (8) holds. Interpreting Eqn. (8) as Ohm’s Law
means the quantity in brackets [ ] represents the total resistance of the circuit. It is called the
impedance and is given the symbol Z. So Ohm’s Law is then: Z I0 = V0. The impedance of this
circuit consists of real and imaginary parts. The real part is the resistance R and the imaginary
part is called the reactance, X. The reactance itself has two parts:
the inductive reactance
the capacitive reactance
XL = w L
XC = 1 / (w C)
(9)
(10)
You can think of inductive and capacitive reactance as A.C. resistances for the inductor and the
capacitor. Hence the impedance can be rewritten as
Z = R + i ( XL – XC )
(11)
Here the complex number Z has been written in the form of: a + ib, where both a and b are real
numbers. This is in direct analogy with 2-dimensional vectors where a is the x component and b
is the y component. Complex numbers can also be written in terms of a magnitude and a phase
factor. This is in direct analogy with 2-dimensional vectors expressed in plane-polar
coordinates.
Z = | Z | e if.
(12)
In this way, | Z | is the length (magnitude) of the vector and phi is the angle measured up from
the positive x axis. This angle is usually written as a theta in plane polar coordinates.
11 – 5
Florida Institute of Technology
Let us examine the real part of the impedance, which is the physically meaningful part.
However, we must develop a few mathematical tools for dealing with complex numbers.
First, the complex conjugate of an imaginary number is called z-star and is written as:
Z* = a – ib
(13)
Then the magnitude of this number (the length of the complex vector) is given by
|Z|
= ( Z Z* )1/2
(14)
Substituting a’s and b’s into Eqn. (14) gives
|Z|
= (a2 + b2 )1/2
Say, Eqn. (15) is just a form of whose Theorem?
And just as with vectors:
(15)
________________________
f = tan-1 (b / a)
(16)
Using Eqns. (13) and (14) on Eqn. (11), we can express the square of the magnitude of the
impedance as
|Z|2
= R2 + ________________________
(17)
Using Eqn. (16) allows us to write:
⎛ Lω − 1

ωC
φ = tan-1 ⎜
R

(18)
Here, phi is interpreted as a phase angle. It determines how the phase of the current differs from
the phase of the input voltage. Referring to Ohm’s Law, the real part of the current can be
written as
I(t) =
V0
cos(ω t − φ )
Z
(19)
In this form, I(t) reaches a maximum when the impedance |Z| is a minimum. When this
happens, the circuit is said to resonate. Eqn. (17), is a minimum when |Z| = R, or when
________________________ = 0
11 – 6
(20)
Florida Institute of Technology
Substitute Eqns. (9) and (10) into Eqn. (20) and rewrite it below.
________________________ = 0
Now solve for the angular frequency w =
(21)
(22)
Resonance occurs only when the driving frequency, w, reaches this value. Resonance in
an RLC circuit can be described in other terms. Notice that at resonance Eqn. (6) indicates that
the impedance equals the total, normal resistance R. In other words at resonance, XC and XL are
equal and cancel one another out. In still other terms, Z = |Z| eif implies that at resonance the
phase factor f = 0. Using tan-1 (0) = 0 in Eqn. (18) gives another way of determining Eqn. (22).
During the experiment, you will notice the input voltage and VR are only in phase at resonance.
During the experiment, you measure VR and compare it with the input voltage V0. By
virtue of Ohm’s Law and R being a constant, VR is always directly proportional to the current.
Hence, resistors are called linear circuit elements. So, the voltage across the resistor is always in
phase with the current flowing in the circuit. By looking at VR on the oscilloscope, you are
equivalently looking at the behavior of the current. The only difference is a scaling factor (the
value of the resistance).
As you perform the experiment, you will find that the current (akin to VR) and the input
voltage are not always in phase with one another. When f > 0, the current reaches a maximum
later than the input voltage (the current lags the voltage), and when f >Approximate cost: \$8 per page.
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