91影视

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

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

$蔚\left(t\right)=\left(45.0V\right)\sin \left({蝇}_{d}t\right)$

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

Formulas:

$Powerfactor=cos蠒$

$P=\frac{1}{2}{蔚}_{m}Icos蠒$

Here, $蠒$ is the phase angle and role="math" localid="1663147875263" ${蔚}_m$ is the emf.

The equation for the power factor is as,

$Powerfactor=cos蠒$

Consider the given equation as below.

$i\left(t\right)=\left(1.20A\right)\sin \left({蝇}_{d}t+42.0掳\right)$ 鈥�.. (1)

Theequation of current is,

$i\left(t\right)=I\sin \left({蝇}_{d}t-蠒\right)$ 鈥�.. (2)

Comparingtheabove equations, so the phase angle will be

$-蠒=42.0掳$or$蠒=-42.0掳$

Thus, the power factor is,

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

Hence, the power factor is 0.743.

As the phase angle is, $蠒<0$.

So,${蝇}_{d}t-蠒$will be positive, that is,${蝇}_{d}t-蠒>{蝇}_{d}t$

Hence, the current leads the emf.

Thephase constant is define by using following equation.

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

Plugging the value of$蠒$,you get

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

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 $蠒=0$, but here $蠒$ is not zero. Hence the circuit is not in resonance.

As the phase angle is,

$蠒=-42.0掳$

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

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

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.