Chapter 21, Problem 93P

In an RLC circuit, these three elements are connected in series: a resistor of 20.0 Ω, a 35.0 mH inductor, and a 50.0 μF capacitor. The ac source of the circuit has an rms voltage of 100.0 V and an angular frequency of 1.0 × 10³ rad/s. (a) Find the rms current and the rms voltage across each of the circuit elements, (b) Does the current lead or lag the source voltage? (c) Draw a phasor diagram, (d) Find the average power dissipated.

To determine

The rms current and rms voltage across each of the circuit element.

Answer to Problem 93P

The rms current across each of the circuit element is $\underset{\_}{4.0\text{A}}$, and voltage across each of the circuit element are $\underset{\_}{V_{R_{rms}}=80\text{V}}$, $\underset{\_}{V_{L_{rms}}=140\text{V}}$, and $\underset{\_}{V_{C_{rms}}=80\text{V}}$.

Explanation of Solution

Write the expression for capacitive reactance.

  $X_C=\frac{1}{\omega C}$        (I)

Here, $\omega$ is the angular frequency, $C$ is the capacitance.

Write the expression for inductive reactance.

  $X_L=\omega L$        (II)

Here, $L$ is the inductance.

Write the expression for impedance.

  $Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$        (III)

Here, $R$ is the resistance.

Write the expression for rms current.

  $I_{rms}=\frac{{\epsilon }_{rms}}{Z}$        (IV)

Write the expression for rms voltage across resistor.

  $V_{R_{rms}}=I_{rms}R$        (V)

Write the expression for rms voltage across inductor.

  $V_{L_{rms}}=I_{rms}{X}_L$        (VI)

Write the expression for rms voltage across capacitor.

  $V_{C_{rms}}=I_{rms}{X}_C$        (VII)

Conclusion:

Substitute, $1.0\times {10}^3\text{rad}/\text{s}$ for $\omega$, $50.0\times {10}^{-6}\text{F}$ for $C$ in equation (I).

  $\begin{array}{c}{X}_C=\frac{1}{\left(1.0\times {10}^3\text{rad}/\text{s}\right)\left(50.0\times {10}^{-6}\text{F}\right)}\\ =20\Omega \end{array}$

Substitute, $1.0\times {10}^3\text{rad}/\text{s}$ for $\omega$, $0.0350\text{H}$ for $L$ in equation (II).

  $\begin{array}{c}{X}_L=\left(1.0\times {10}^3\text{rad}/\text{s}\right)\left(0.0350\text{H}\right)\\ =35\Omega \end{array}$

Substitute, $20\Omega$ for $R$, $20\Omega$ for $X_C$, and $35\Omega$ for $X_L$ in equation (III).

  $\begin{array}{c}Z=\sqrt{{\left(20\Omega \right)}^2+{\left(35\Omega -20\Omega \right)}^2}\\ =25\Omega \end{array}$

Substitute, $100\text{V}$ for ${\epsilon }_{rms}$, $25\Omega$ for $Z$ in equation (IV).

  $\begin{array}{c}{I}_{rms}=\frac{100\text{V}}{25\Omega }\\ =4.0\text{A}\end{array}$

Substitute, $20\Omega$ for $R$, $4.0\text{A}$ for $I_{rms}$ in equation (V) to find $V_{R_{rms}}$.

  $\begin{array}{c}{V}_{R_{rms}}=\left(4.0\text{A}\right)\left(20\Omega \right)\\ =80\text{V}\end{array}$

Substitute, $35\Omega$ for $X_L$, $4.0\text{A}$ for $I_{rms}$ in equation (VI) to find $V_{L_{rms}}$.

  $\begin{array}{c}{V}_{L_{rms}}=\left(4.0\text{A}\right)\left(35\Omega \right)\\ =140\text{V}\end{array}$

Substitute, $20\Omega$ for $X_C$, $4.0\text{A}$ for $I_{rms}$ in equation (VI) to find $V_{L_{rms}}$.

  $\begin{array}{c}{V}_{C_{rms}}=\left(4.0\text{A}\right)\left(20\Omega \right)\\ =80\text{V}\end{array}$

Therefore, the average current through the coil during rotation is $\underset{\_}{0.57\text{A}}$.

To determine

Whether the current lead or lag voltage.

Answer to Problem 93P

Current lags voltage.

Explanation of Solution

The value of inductive reactance is $35\Omega$, and the value of capacitive reactance is $20\Omega$. Since $X_L>X_C$, inductor dominates capacitor. Thus the current lags voltage.

Conclusion:

Therefore, Current lags voltage.

To determine

Sketch the phasor diagram.

Answer to Problem 93P

The phasor diagram is shown below.

Explanation of Solution

Write the expression for phase angle.

  $\varphi ={\cos }^{-1}\left(\frac{R}{Z}\right)$        (VIII)

Conclusion:

Substitute, $20\Omega$ for $R$, and $25\Omega$ for $Z$ in equation (VIII).

  $\begin{array}{c}\varphi ={\cos }^{-1}\left(\frac{20\Omega }{25\Omega }\right)\\ =37°\end{array}$

Therefore, the phasor diagram is.

To determine

The average power dissipated.

Answer to Problem 93P

The average power dissipated is $\underset{\_}{320\text{W}}$.

Explanation of Solution

Write the expression for power dissipated.

  $P=I^2{}_{rms}R$        (IX)

Conclusion:

Substitute, $20\Omega$ for $R$, and $4.0\text{A}$ for $I_{rms}$ in equation (IX).

  $\begin{array}{c}P={\left(4.0\text{A}\right)}^2\left(20\Omega \right)\\ =320\text{W}\end{array}$

Therefore, the average power dissipated is $\underset{\_}{320\text{W}}$.