Physics 102 Alternate Current RCL Circuit Lab Key
Predictions
1. Based on the equations XC= 1/( 2πfC) and XL = 2πfL,
a. Is the capacitive reactance at maximum at high or low frequencies?
Low
b. Is the inductive reactance at maximum at high or low frequencies?
High
2. Based on what you learned from Chapter 23 in your text book, at resonance frequency,
a. Is the current at maximum or minimum?
Max
b. Is the total impedance of the circuit at maximum or minimum?
Min
Problem Solving
1. A series RCL circuit has a resonant frequency of 1500 Hz. When operating at a frequency other than 1500 Hz,
the circuit ha a capacitive reactance of 5 Ω and an inductive reactance of 30 Ω. What are the values of L and C?
REASONING Since the resonant frequency f0 is known, we may use the equation, f 0 
1
2  LC
to
find the inductance L, provided the capacitance C can be determined. The capacitance can be found by
using the definitions of capacitive and inductive reactances.
SOLUTION
a. Solving for the inductance, we have L 
1
4  f 02 C
(1)
2
where f0 is the resonant frequency. From the equations in your book, the capacitive and inductive
reactances are X C 
1
2 f C
and
X L  2 f L
where f is any frequency. Solving the first of these equations for f, substituting the result into the
second equation, and solving for C yields C 
L
. Substituting this result into Equation (1)
XL XC
above and solving for L gives
L 
1
2 f 0
XL XC 
1
2  1500 Hz
b
g b 30.0  gb5.0  g 
1.3  10 3 H
b. The capacitance is C 
1.3  10 3 H
L

 8.7  10 6 F
XL XC
30.0  5.0 
b
gb
g
2. A series RCL circuit includes a resistance of 275 Ω, and inductive reactance of 648 Ω, and a capacitive
reactance of 415 Ω. The current in the circuit is 0.233 A. What is the voltage of the generator?
REASONING The voltage supplied by the generator can be found from the equation: Vrms  I rms Z .
The value of I rms is given in the problem statement, so we must obtain the impedance of the circuit.
SOLUTION The impedance of the circuit is, according to Equation 23.7,
Z
R 2  ( X L – X C ) 2  ( 275  ) 2  (648  – 415  ) 2  3.60  10 2 
The rms voltage of the generator is Vrms  I rms Z  ( 0.233 A )( 3.60  10 2  ) = 83.9 V
3. The elements in a series RCL circuit are a 106 Ω resistor, a 3.30 μF capacitor, and a 0.0310 H inductor. The
frequency is 609 Hz. What are (a) the impedance of the circuit (b) the phase angle between the current and the
voltage of the generator (c) the average power consumed of the circuit if it is plugged into a standard household
power outlet (120 V)?
REASONING AND SOLUTION
a. At a frequency of 609 Hz:
XL = 2 fL = 2(609 Hz)(0.0310 H) = 119 
XC 
1
1

2 f C
2  609 Hz 3.30  10 6 F
b
gc
so,
Z 
c
R2  X L  XC
h
2

h
 79 .2 
b106 g  b119   79.2 g
2
2
 113 
b. The phase angle between the current and the voltage is
  tan 1
FG X
H
L
 XC
R
IJ
K
 tan 1
FG 119   79.2  IJ
H 106  K

21
c. The Power can be found from this equation: P = IVcosφ and I = V/Z, so P = V2/Z (cosφ)
P = (120)2/(113.1) [cos(20.40)] = 119.33 W