Problem + Hard 53-3    Source:   2nd question - page 10 - BORDINI, Sergio Machado -   Book: Circuit: Solving Problems - 1st Edition - Manuscript (621.3.049.B729c, Eng. Ufrgs Library).

In the circuit shown in the figure below we have an impedance Z of unknown nature. To find it, two readings were taken on the voltmeters where V₁ = 30 volts and V₂ = 90 volts. It is known that I₁ is ahead of a E₁. Determine the impedance Z.

Observation - In his book the author presented the solution of the problem using the graphical method. In this case, we will use the analytical method for the solution.

Solution of the Problem + Hard 53-3

As the problem statement informs that the current I₁ is leading in relation to a E₁ then we conclude that there will be a capacitive predominance in the impedance Z. If Z is a pure capacitor then the phasor sum of the two multimeters should equal 100 volts. Checking: 30² + 90² = 9,000; and this should equal 100² = 10,000. As it is not, we conclude that Z is NOT a pure capacitor. Surely there must be a resistor in series with the capacitor. For a better understanding of the problem, we built a circuit that shows the voltages on the components as we can see in the figure below.

According to the figure above, we can write the equation that allows us to find the solution to the problem. Remember that the phasor sum of the voltages across the resistors and the capacitor must equal the source voltage. Thus, squaring the parcels we find:

From this equation we easily find the value of V_R, or:

With this value in hand, we can find the value of the voltage across the capacitor, or:

Substituting for numerical values ​​we find:

On the other hand, since we have the value of the voltage across the 10 ohms resistor, we can calculate the value in I₁.

With this result we can calculate the values ​​of R and X_C. Then:

And the value of X_C is:

Therefore the impedance Z can be represented by:

To know how far the current I₁ is ahead of E₁ we must find the equivalent impedance that the circuit offers to the voltage source, that is:

Now just find the quotient between the source voltage and the equivalent impedance, or:

Confirming the measurement made by the voltmeter V₁ (3 x 10 = 30 V) and the statement that the current is leading (by 62.17°) in relation to the voltage applied to the circuit.