Problem + Hard 52-2
Source: Adapted from Problem 39 - List of Problems Circuits II - Discipline Electric Circuits of
School of Engineering - UFRGS - 2017 - Prof. Dr. Valner Brusamarello.
For the circuit shown in the figure below, it is known that wattmeter 1, W1, has a reading equal to
at 20 kW. Determine:
a) The value of E
b) The wattmeter 2 reading, W2.
Solution of the Problem + Hard 52-2
Item a
Initially, we will calculate the value of I1 as a function of E, knowing that the
equivalent impedance of the branch is Zeq1= 1 + 0.5j + 0.5j = 1 + j = √2∠45°. Then:
I1 = E / (1 + j) = E∠-45° / √2
For the current I2 the equivalent branch impedance is Zeq2 = -j - j3 + j2 + 2 = 2 - j2 =
2√2∠-45°. Soon:
I2 = E / (2 - j2) = E∠45° / 2√2
And finally, to calculate I3 we use the equivalent impedance of the corresponding branch which is
Zeq3 = 4 + j - j = 4. Like this:
I3 = E / 4
As the value of the real power measured by the wattmeter 1 was provided, we need to find the value
of V1, this value given by the difference Vad - Vbd.
Then:
Vad = 0.5∠90° x I1 = 0.5∠90° x E∠-45° / √2
Carrying out the calculation:
Vad = E∠45° / 2√2 = E (0.25 + j0.25)
Now let's calculate the value of Vbd. Based on the initial circuit we have that
Zbd = -j3 + j2 + 2 = 2 - j = √5∠-26.57°. Then:
Vbd = Zbd. I2 = √5∠-26.57° x E∠45° / 2√2
Carrying out the calculation:
Vbd = E∠18.43° (√5 / 2√2) = E (0.75 + j0.25)
We are now ready to calculate the value of V1 = Vad - Vbd.
V1 = E [(0.25 + j0.25) - (0.75 + j0.25)]
As only the module of V1 is of interest, carrying out the calculation:
|V1| = 0.5 E
By the circuit, the power measured by the wattmeter is the real part of the product between
|V1| and |I2|. Soon:
W1 = 20,000 = 0.5 x E (E / √2) x cos (-45°) = E2 / 4
So the value of E is:
E = √(4 W1) = √(80,000) = 200√2 volts
Item b
Let's calculate the value of I3 knowing that Zeq3 = 3 - j + 1 + j = 4. Then:
I3 = E / 4 = 200√2 / 4 = 50√2 A
On the other hand, we know that the power measured by the wattmeter 2 will be the real part of the product
between |V2| and |I3|. To calculate |V2| we must find the values of
Vcd and Ved because |V2| = Vcd - Ved.
The value of I2 has already been calculated as a function of E. Thus, replacing E by its value, we obtain: we have:
I2 = E∠45° / 2√2 = 200√2 / 2√2 = 100∠45° A
Now we can calculate the value of Vcd, or:
Vcd = 2 I2 = 200∠45° V
To calculate the value of Ved, it is known that Zed = 1 + j = √2∠45°< /x>. Soon:
