Problem + Hard 52-1
Source: Problem 46 - List of Problems Circuits II - Discipline
Electrical Circuits of the School of Engineering - UFRGS - 2017 - Prof. Dr. Valner Brusamarello.
Knowing that the wattmeter reading is 7.5 kW, determine the value of E.
Solution of the Problem + Hard 52-1
Initially notice that the current source and the circuit (between points a and c) formed by the resistance
of 4 ohms in series with the inductor j4, can be eliminated from the circuit as they are in parallel with
the voltage source.
In this way, the circuit reduces to the one shown in the figure below.
In this circuit we perform the calculation of the series circuits and so we can calculate the value of
I1 and I2 as a function of E. Soon:
I1 = E / (2 + j2) = E∠-45° / 2√2
For the current I2, we have:
I2 = E / -j4 = E∠90° / 4
Remember that j4 = 4∠-90°. In possession of these values and returning to the initial circuit
we can calculate the voltage between points k and c, or:
Vkc = j3 I1 = 3∠90° (E∠-45° / 2√2)
Doing the calculation we find:
Vkc = 3(E∠45° / 2√2)= E(1.061 cos 45°+ j1.061 sen 45°)
Now let's calculate the value of Vmc. Based on the initial circuit we have:
Vmc = -j4 I2 = 4∠-90°. (E∠90° / 4) = E
From the problem we know that the measurement of the voltmeter V = Vkm = Vkc - Vmc. So let's find that value.
Vkm = E(1.061 cos 45°+ j1.061 sen 45°) - E
Carrying out the calculation:
Vkm = E (-0.25 + j0.75) = 0.79 x E∠108.44°
By the circuit, the power measured by the wattmeter is the real part of the product between Vkm and I2.
Soon:
P = 7 500 = 0.79 x E∠108.44° x E∠90° / 4
Rearranging the terms and using the cosine of the angle, as we are only interested in the real part, we find:
E = √(7 500/ |0,1975 cos 198,44°|) = √(40 030)
Taking the square root we find the value of E, or:
