Figure 31-36 shows an ac generator connected to a “black box” through a pair of terminals. The box contains an RLC circuit, possibly even a multiloop circuit, whose elements and connections we do not know. Measurements outside the box reveal that$\mathit{\epsilon }\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathbf{75}\mathbf{.}\mathbf{0}\mathbf{V}\mathbf{)}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}{\mathbf{\omega }}_{\mathbf{d}}\mathbf{t}\mathbf{\right)}$and $\mathit{i}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{A}\mathbf{)}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}{\mathbf{\omega }}_{\mathbf{d}}\mathbf{t}\mathbf{+}\mathbf{42}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{)}$.

Figure 31-36 is the circuit diagram with the unknown element.

$i\left(t\right)=\left(1.20A\right)\sin \left({\omega }_{d}t+42.0°\right)$

$\epsilon \left(t\right)=\left(45.0V\right)\sin \left({\omega }_{d}t\right)$

Use the concept of power factor, resonance, and power. Using the equations, determine and find the required values.

Formulas:

$Powerfactor=cos\varphi$

$P=\frac{1}{2}{\epsilon }_{m}Icos\varphi$

Here, $\varphi$ is the phase angle and role="math" localid="1663147875263" ${\epsilon }_m$ is the emf.

The equation for the power factor is as,

$Powerfactor=cos\varphi$

Consider the given equation as below.

$i\left(t\right)=\left(1.20A\right)\sin \left({\omega }_{d}t+42.0°\right)$ ….. (1)

Theequation of current is,

$i\left(t\right)=I\sin \left({\omega }_{d}t-\varphi \right)$ ….. (2)

Comparingtheabove equations, so the phase angle will be

$-\varphi =42.0°$or$\varphi =-42.0°$

Thus, the power factor is,

$$\begin{gathered} Powerfactor=\text{cos}\left(-42.0°\right) \\ =0.743 \end{gathered}$$

Hence, the power factor is 0.743.

As the phase angle is, $\varphi <0$.

So,${\omega }_{d}t-\varphi$will be positive, that is,${\omega }_{d}t-\varphi >{\omega }_{d}t$

Hence, the current leads the emf.

Thephase constant is define by using following equation.

$\tan \varphi =\frac{\left(X_L-X_C\right)}{R}$

Plugging the value of$\varphi$,you get

$$\begin{gathered} \tan \left(-42.0°\right)=\frac{\left(X_L-X_C\right)}{R} \\ \frac{\left(X_L-X_C\right)}{R}=-0.900 \\ \left(X_L-X_C\right)=-0.900R \end{gathered}$$

From this equation, we get the value of$\left(X_L-X_C\right)$as negative, means$\left(X_L<X_C\right)$.

So, the circuit in the box is capacitive.

For resonance, $X_L=X_C$, and therefore, the angle $\varphi =0$, but here $\varphi$ is not zero. Hence the circuit is not in resonance.

As the phase angle is,

$\varphi =-42.0°$

It is negative and finite, and the $X_C$ is not zero. So, you can say that the box must contain capacitor.

As $\left(X_L<X_C\right)$.

From this relation, we can determine that $X_L$ may be zero, so there may not be an inductor in the box.

Must there be a resistor in the box. The circuit is RLC, and R is not zero in the box, so there must be a resistor in the box.

We can use equation

$$\begin{gathered} P_{avg}=\frac{1}{2}{\epsilon }_{m}cos\varphi \\ =\frac{1}{2}\left(75.0\right)\left(1.20\right)\left(0.743\right) \\ =33.4W \end{gathered}$$

Hence, the average rate at which energy is delivered to the box by the generator is $P_{avg}=33.4W$

Because all the answers depend only on the phase angle.