Chapter 21, Problem 95P

(a)

To determine

The magnitude of impedance of the circuit.

Answer to Problem 95P

The impedance of the circuit is $13.5\Omega$.

Explanation of Solution

Write the expression to calculate the impedance of the circuit.

  $Z=\sqrt{R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2}$

Write the expression to calculate $\omega$.

  $\omega =2\pi f$

Here, f is the frequency.

Rewrite the equation for Z using the above equation.

  $Z=\sqrt{R^2+{\left(2\pi fL-\frac{1}{2\pi fC}\right)}^2}$

Conclusion:

Substitute $12.0\Omega$ for R, $2.50\text{kHz}$ for f, $0.26\text{μF}$ for C and $15.2\text{mH}$ for L in the above equation to calculate Z.

  $\begin{array}{c}Z=\sqrt{{\left(12.0\Omega \right)}^2+{\left(\begin{array}{l}2\pi \left(2.50\text{kHz}\right)\left(\frac{{10}^3\text{Hz}}{1\text{kHz}}\right)15.2\text{mH}\left(\frac{{10}^{-3}\text{H}}{1\text{mH}}\right)\\ -\left(\frac{1}{2\pi \left(2.50\text{kHz}\right)\left(\frac{{10}^3\text{Hz}}{1\text{kHz}}\right)0.26\text{μF}\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)}\right)\end{array}\right)}^2}\\ =\sqrt{144{\Omega }^2+{\left(238.76{\Omega }^2-244.85{\Omega }^2\right)}^2}\\ =13.5\Omega \end{array}$

(b)

To determine

The magnitude of rms value of current.

Answer to Problem 95P

The rms value of current is $\text{18}\text{A}$.

Explanation of Solution

Write the expression to calculate the rms current.

  $I_{\text{rms}}=\frac{V_{\text{rms}}}{Z}$

Here, $I_{\text{rms}}$ is the rms value of current and $V_{\text{rms}}$ is the rms value of voltage or power supply.

Conclusion:

Substitute $240\text{V}$ for $V_{\text{rms}}$ and $13.5\Omega$ for Z in the above equation to calculate $I_{\text{rms}}$.

  $\begin{array}{c}{I}_{\text{rms}}=\frac{240\text{V}}{13.5\Omega }\\ =17.7\text{A}\\ \approx \text{18}\text{A}\end{array}$

(c)

To determine

The magnitude of phase angle.

Answer to Problem 95P

The phase angle is $-27.3°$.

Explanation of Solution

Write the expression to calculate the phase angle.

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

Here, $\varphi$ is the phase angle.

Conclusion:

Substitute $12.0\Omega$ for R and $13.5\Omega$ for Z in the above equation to calculate $\varphi$.

  $\begin{array}{c}\varphi ={\cos }^{-1}\left(\frac{12.0\Omega }{13.5\Omega }\right)\\ =27.3°\end{array}$

Since the inductive reactance is less than capacitive reactance, the phase angle is negative. Thus, the phase angle is $-27.3°$.

(d)

To determine

Whether the current lead or lag the voltage.

Answer to Problem 95P

The rms value of current is $\text{18}\text{A}$.

Explanation of Solution

Write the expression to calculate the ratio of capacitive reactance to the inductive reactance.

  $r=\frac{\left(\frac{1}{\omega C}\right)}{\omega L}$

Here, r is the ratio.

Rewrite the above expression using the expression for $\omega$.

  $\begin{array}{c}r=\frac{\left(\frac{1}{2\pi fC}\right)}{2\pi fL}\\ =\frac{1}{4{\pi }^2{f}^{2}LC}\end{array}$

Substitute $2.50\text{kHz}$ for f, $0.26\text{μF}$ for C and $15.2\text{mH}$ for L in the above equation to calculate r.

  $\begin{array}{c}r=\frac{\left(\frac{1}{2\pi fC}\right)}{2\pi fL}\\ =\frac{1}{4{\pi }^2{\left(2.50\text{kHz}\left(\frac{{10}^3\text{Hz}}{1\text{kHz}}\right)\right)}^2\left(15.2\text{mH}\left(\frac{{10}^{-3}\text{H}}{1\text{mH}}\right)\right)0.26\text{μF}\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)}\\ =1.02\end{array}$

Since the ratio is greater than 1, the value of capacitive reactance is greater than the inductive reactance. Thus, the current leads the voltage.

Conclusion:

(e)

To determine

The magnitude of rms voltages across the elements.

Answer to Problem 95P

The rms voltage across the resistor, inductor and capacitor is respectively $216\text{V}$, $\text{4}\text{.3}\times {\text{10}}^3\text{V}$ and $4.4\times {10}^3\text{V}$.

Explanation of Solution

Write the expression to calculate the voltage across the resistor.

  $V_R=I_{rms}R$

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

Substitute $12.0\Omega$ for R and $\text{18}\text{A}$ for $I_{rms}$ in the above equation to calculate $V_R$.

  $\begin{array}{c}{V}_R=\left(12.0\Omega \right)\text{18}\text{A}\\ =216\text{V}\end{array}$

Write the expression to calculate the voltage across the inductor.

  $V_L=I_{rms}\omega L$

Here, $V_L$ is the voltage across the inductor.

Rewrite the above equation using the expression for $\omega$.

  $V_L=I_{rms}\left(2\pi f\right)L$

  Substitute $\text{18}\text{A}$ for $I_{rms}$, $2.50\text{kHz}$ for f and $15.2\text{mH}$ for L in the above equation to calculate $V_L$

  $\begin{array}{c}{V}_L=\text{18}\text{A}\left(2\pi \left(2.50\text{kHz}\left(\frac{{10}^3\text{Hz}}{1\text{kHz}}\right)\right)\right)15.2\text{mH}\left(\frac{{10}^{-3}\text{H}}{1\text{mH}}\right)\\ =4.29\times {10}^3\text{V}\\ \approx \text{4}\text{.3}\times {\text{10}}^3\text{V}\end{array}$

Write the expression to calculate the voltage across the capacitor.

  $V_C=\frac{I_{rms}}{\omega C}$

Rewrite the above equation using the expression for $\omega$.

  $V_C=\frac{I_{rms}}{2\pi fC}$

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

  Substitute $\text{18}\text{A}$ for $I_{rms}$, $2.50\text{kHz}$ for f and $0.26\text{μF}$ for C in the above equation to calculate $V_C$

  $\begin{array}{c}{V}_C=\frac{\text{18}\text{A}}{2\pi \left(2.50\text{kHz}\left(\frac{{10}^3\text{Hz}}{1\text{kHz}}\right)\right)\left(0.26\text{μF}\right)\left(\frac{{10}^{-6}\mathrm{F}}{1\text{μF}}\right)}\\ =4.4\times {10}^3\text{V}\end{array}$

Conclusion:

Therefore, the rms voltage across the resistor, inductor and capacitor is respectively $216\text{V}$, $\text{4}\text{.3}\times {\text{10}}^3\text{V}$ and $4.4\times {10}^3\text{V}$.