Problem 58-8
Source: Problem created by the site's authors.
For the circuit shown in Figure 58-08.1 calculate:
a) the value of Z so that the circuit is in resonance at a frequency of 100 Hz;
c) calculate the values of L and C;
b) and the circuit bandwidth.
Figure 58-08.1
Solution of the Problem 58-8
Item a
We know that for a circuit to be in resonance condition, the impedance
total of the circuit must be a real value. To solve this problem we will calculate the impedances between the points a-m and b-m. How do the values
of the components are the same so the results will be the same. Calculating the parallel, we obtain
And we know that these reactances are connected in series. Therefore, the equivalent reactance, Xeq, is:
Xeq = - j20 + (- j 20) = - j 40 Ω
On the other hand, the 35 Ω and 85 Ω resistors are in series. Therefore, it results in a
120 Ω resistor. This resistor is in parallel with the 60 Ω resistor. Calculating the parallel
of these resistors we find an equivalent resistor Req of:
Req = 60 x 120 / (60 + 120) = 40 Ω
And since Req and Xeq are connected in parallel, solving this parallel
we find the equivalent impedance, Zeq, between points a - b. Soon
Zeq = - j 40 (40) / ( 40 - j40) = 20 - j 20
Therefore, as the impedance
the total value of the circuit must be a real value, so we can easily find the value of Z, or:
Z = + j 20
In this case, we have a RLC series circuit.
Item b
As we know, the inductive reactance is given by XL = 2 π f L. So the value of L is:
L = 20 / 2 π 100 = 31.83 mH
On the other hand, the definition of capacitive reactance is XC = 1 / 2 π f C. So the value of C is:
C = 1 / 2 π 100 20 = 79.60 µF
Item c
Bandwidth can be calculated using the equation reproduced below.
eq. 58-06
Replacing the variables with their respective numerical values and remembering that
R = 20 Ω, we have:
Δ f = 20 / (2 x π x 31.83 x 10-3) = 5 Hz
Note that if we increase the value of R the bandwidth increases and vice versa.
