Chapter 21, Problem 59P

(a)

To determine

The angular resonant frequency.

Answer to Problem 59P

The angular resonant frequency is $\underset{\_}{745\text{rad/s}}$.

Explanation of Solution

The diagram for the RLC circuit is given figure 1:

  

Write the equation of resonant frequency.

  ${\omega }_0=\sqrt{\frac{1}{LC}}$                                                                                              (I)

Here, ${\omega }_0$ is the resonant frequency, $L$ is the inductance and $C$ is the capacitance.

Conclusion:

Substitute $0.300\text{H}$ for $L$ and $6.00\text{μF}$ for $C$ in equation (I) to find ${\omega }_0$.

  $\begin{array}{c}{\omega }_0=\sqrt{\frac{1}{\left(0.300\text{H}\right)\left[\left(6.00\text{μF}\right)\left(\frac{1\times {10}^{-6}\text{F}}{1\text{μF}}\right)\right]}}\\ =\sqrt{\frac{1}{\left(0.300\text{H}\right)\left(6.00\times {10}^{-6}\text{F}\right)}}\\ =745\text{rad/s}\end{array}$

Thus, the angular resonant frequency is $\underset{\_}{745\text{rad/s}}$.

(b)

To determine

The value of resistance in the circuit.

Answer to Problem 59P

The value of resistance in the circuit is $\underset{\_}{790\Omega }$.

Explanation of Solution

Write the equation of resistance.

  $R=\frac{{\epsilon }_m}{I}$                                                                                                               (II)

Here, $R$ is the resistance, ${\epsilon }_m$ is the peak voltage and $I$ is the current.

Conclusion:

Substitute $440\text{V}$ for ${\epsilon }_m$ and $\text{0}\text{.560}\text{A}$ for $I$ in equation (II) to find $R$.

  $\begin{array}{c}R=\frac{\left(440\text{V}\right)}{\left(\text{0}\text{.560}\text{A}\right)}\\ =790\Omega \end{array}$

Thus, the value of resistance in the circuit is $\underset{\_}{790\Omega }$.

(c)

To determine

The peak voltage across the resistor, inductor and capacitor at resonant frequency.

Answer to Problem 59P

The peak voltage across the resistor, inductor and capacitor at resonant frequency are respectively $\underset{\_}{440\text{V},125\text{V}\text{and}125\text{V}}$.

Explanation of Solution

Write the equation for the peak voltage across resistor.

  $V_R={\epsilon }_m$                                                                                                             (III)

Here, $V_R$ is the peak voltage across resistor.

Write the equation of voltage across inductor.

  $V_L=I{X}_L$                                                                                                            (IV)

Here, $V_L$ is the voltage across inductor, $I$ is the current and $X_L$ is the inductive reactance.

Write the equation of the inductive reactance.

  $X_L={\omega }_{o}L$                                                                                                        (V)

Here, $L$ is the inductance.

Rewrite the expression for the peak voltage across inductor by using equation (I), (IV) and (V).

  $V_L=I\sqrt{\frac{L}{C}}$                                                                                                         (VI)

At resonant frequency, the voltage across inductor and capacitor will be same.

Write the equation for the peak voltage across capacitor.

  $V_C=V_L$                                                                                                            (VII)

Here, $V_C$ is the peak voltage across capacitor.

Conclusion:

Substitute $440\text{V}$ for ${\epsilon }_m$ in equation (III) to find $V_R$.

  $V_R=440\text{V}$

Substitute $0.300\text{H}$ for $L$, $0.560\text{A}$ for $I$ and $6.00\text{μF}$ for $C$ in equation (VI) to find $V_L$.

  $\begin{array}{c}{V}_L=\left(0.560\text{A}\right)\sqrt{\frac{\left(\left(0.300\text{H}\right)\right)}{\left[\left(6.00\text{μF}\right)\left(\frac{1\times {10}^{-6}\text{F}}{1\text{μF}}\right)\right]}}\\ =\left(0.560\text{A}\right)\sqrt{\frac{\left(\left(0.300\text{H}\right)\right)}{\left(6.00\times {10}^{-6}\text{F}\right)}}\\ =125\text{V}\end{array}$

Substitute $125\text{V}$ for $V_L$ in equation (VII) to find $V_C$.

  $V_C=125\text{V}$

Therefore, the peak voltage across the resistor, inductor and capacitor at resonant frequency are respectively $\underset{\_}{440\text{V},125\text{V}\text{and}125\text{V}}$.