Problem 52-3
Source: Problem 11 - page 448 -
HAYT, William Jr, KEMMERLY, Jack E. , DURBIN, Steven M. - Book: Análise de Circuitos em Engenharia - 7ª edição - Ed. McGraw Hill - 2008.
Determine the average power absorbed by each of the circuit elements
shown in the Figure 52-03.
Figure 52-03
Solution of the Problem 52-3
Since the inductor is in parallel with the series circuit formed by the capacitor and the resistance of
10 ohms, we can calculate the equivalent impedance of this association, highlighted in green. The resistance series can be represented with the capacitor as 10 - j5 = 11.18 ∠-26.56° and the inductor as j5 = 5 ∠90°, then
Zeq = 11.18 ∠-26.56° x 5 ∠90° / (10 - j5 + j5 )
Performing the calculation:
Zeq = 55.90 ∠+63.43° / 10 = 5.59 ∠+63.43°
Or, in rectangular form:
Zeq = 2.5 + j5 Ω
Therefore, the circuit highlighted in green can be replaced by a resistor of
2.5 ohms in series with an inductor of jXL = j5 ohms. Adding this impedance to the resistance of 4 ohms, we find the total impedance that the circuit offers to the voltage source, or:
Knowing the value of the total impedance can calculate the value of the current
IT that circulates through the circuit.
IT = V / Ztotal = 100 ∠0° / 8.20 ∠+37.57°
When the calculation is made, the value of IT is:
IT = 12.20 ∠-37.57° A
As the current IT is delayed in relation to the voltage, then the
circuit has inductive predominance and the power factor of the circuit is:
PF = cos ∠-37.57° = 0.79 indutivo
Knowing the value of IT we can calculate the values of
IC and IL in two different ways.
Alternative 1
Knowing the values of IT and Zeq
there are conditions to determine the potential difference between points a - b, and from this data, we calculate IC and IL.
Vab = Zeq IT = 5.59 ∠+63.43° x 12.20 ∠-37.57°
By performing the calculation, the value of Vab is:
Vab = 68.20 ∠+25.86° volts
Applying the Ohm's law, we calculate IL, or:
IL = Vab / 5 ∠+90° = 68.20 ∠+25.86° / 5 ∠+90°
IL = 13.64 ∠-64.14° A
For the calculation of IC, applies to Ohm's law, or:
IC = Vab / 11.18 ∠-26.56°
Note that the impedance of the RC circuit in the polar form was used,
where 10 - j5 = 11.18 ∠-26.56°. So:
IC = 68.20 ∠+25.87° / 11.18 ∠-26.56° = 6.10 ∠+52.43° A
Alternative 2
The other way to calculate IC and IL is using a
current divider.
IL = (IT Zeq ) / 5 ∠+90°
IL = (12.20 ∠-37.57° x 5.59 ∠+63.43°) / 5 ∠+90°
IL = 13.64 ∠-64.14° A
And the value of IC is:
IC = (IT Zeq ) / 11.18 ∠-26.56°
IC = (12.20 ∠-37.57° x 5.59 ∠+63.43°) / 11.18 ∠-26.56°
IC = 6.10 ∠+52.43° A
We have all the data needed to calculate the powers. Initially, we will calculate the powers in the resistors.
P4 = R4 |IT|2 = 4 x 12.202 = 595.36 W
P10 = R10 |IC|2 = 10 x 6.102 = 372.10 W
We known that resistors dissipate average or real power. Adding the two values
found, the total average power supplied by the voltage source is:
PS = P4 + P10 = 595.36 + 372.10 = 967.46 W
To supplement the problem, let's calculate the other types of powers. In the capacitor and the inductor there is only reactive power, represented by the letter Q.
Recalling that in the capacitor the reactive power is NEGATIVE and in the inductor POSITIVE.
QC = - XC |IC|2 = - 5 x 6.102 = - 186.05 VAr
QL = XL |IL|2 = 5 x 13.642 = 930.25 VAr
Therefore, the total reactive power supplied by the voltage source is:
Qtotal = QL + QC = 930.25 - 186.05 = 744.2 VAr
To confirm the results we can calculate the apparent power provided by the voltage source in two ways, that is:
Stotal = √ (Ptotal2 + Qtotal2)
Substituting for the numerical values, we find S, or:
Stotal = √ (967.462 + 744.202) = 1 220.58 VA
On the other hand, we can calculate the value of apparent power, S by the following expression:
Stotal = |V| |IT| = 100 x 12.20 = 1 220 VA
We found a difference of 0.58 VA by virtue of rounding.
