Chapter 33, Problem 33.59AP

Review. The voltage phasor diagram for a certain series RLC circuit is shown in Figure P33.59. The resistance of the circuit is 75.0 Ω, and the frequency is 60.0 Hz. Find (a) the maximum voltage ΔV_max, (b) the phase angle ϕ, (c) the maximum current, (d) the impedance, (e) the capacitance and (f) the inductance of the circuit, and (g) the average power delivered to the circuit.

To determine

 The maximum voltage.

Answer to Problem 33.59AP

The maximum voltage is $22.4\text{V}$.

Explanation of Solution

Given info: The resistance of the circuit is $75.0\Omega$, the frequency is $60.0\text{Hz}$. The voltage drop across the resistor is $20.0\text{V}$, the voltage drop across the inductor is $25.0\text{V}$ and the voltage drop across the capacitor is $15.0\text{V}$.

The expression for maximum value of the voltage is,

$\Delta {\text{V}}_{\text{max}}=\sqrt{{\left(\Delta {\text{V}}_{\text{R}}\right)}^2+{\left(\Delta {\text{V}}_{\text{L}}-\Delta {\text{V}}_{\text{C}}\right)}^2}$

Here,

$\Delta {\text{V}}_{\text{R}}$ is the voltage drop across the resistor.

$\Delta {\text{V}}_{\text{L}}$ is the voltage drop across the inductor.

$\Delta {\text{V}}_{\text{C}}$ is the voltage drop across the capacitor.

Substitute $20.0\text{V}$ for $\Delta {\text{V}}_{\text{R}}$, $25.0\text{V}$ for $\Delta {\text{V}}_{\text{L}}$ and $15.0\text{V}$ for $\Delta {\text{V}}_{\text{C}}$ in the above expression.

$\begin{array}{c}\Delta {\text{V}}_{\text{max}}=\sqrt{{\left(20.0\text{V}\right)}^2+{\left(25.0\text{V}-15.0\text{V}\right)}^2}\\ =22.36\text{V}\\ \approx 22.4\text{V}\end{array}$

Conclusion:

Therefore, the maximum voltage is $22.4\text{V}$.

To determine

 The phase angle.

Answer to Problem 33.59AP

 The phase angle is $26.6°$.

Explanation of Solution

Given info: The resistance of the circuit is $75.0\Omega$, the frequency is $60.0\text{Hz}$. The voltage drop across the resistor is $20.0\text{V}$, the voltage drop across the inductor is $25.0\text{V}$ and the voltage drop across the capacitor is $15.0\text{V}$.

The expression for the phase angle is,

$\varphi ={\sin }^{-1}\left(\frac{\Delta {\text{V}}_{\text{L}}-\Delta {\text{V}}_{\text{C}}}{\Delta {\text{V}}_{\text{max}}}\right)$

Substitute $22.4\text{V}$ for $\Delta {\text{V}}_{\text{max}}$, $25.0\text{V}$ for $\Delta {\text{V}}_{\text{L}}$ and $15.0\text{V}$ for $\Delta {\text{V}}_{\text{C}}$ in the above expression.

$\begin{array}{c}\varphi ={\sin }^{-1}\left(\frac{25.0\text{V}-15.0\text{V}}{22.4\text{V}}\right)\\ =26.5°\\ \approx 26.6°\end{array}$

Conclusion:

Therefore, the phase angle is $26.6°$.

To determine

 The maximum current.

Answer to Problem 33.59AP

The maximum current is $0.267\text{A}$.

Explanation of Solution

Given info: The resistance of the circuit is $75.0\Omega$, the frequency is $60.0\text{Hz}$. The voltage drop across the resistor is $20.0\text{V}$, the voltage drop across the inductor is $25.0\text{V}$ and the voltage drop across the capacitor is $15.0\text{V}$.

The expression for maximum current is,

$I_{\text{max}}=\frac{\Delta {\text{V}}_{\text{R}}}{R}$

Here,

$R$ is the resistance.

Substitute $20.0\text{V}$ for $\Delta {\text{V}}_{\text{R}}$ and $75.0\Omega$ for $R$ in the above expression.

$\begin{array}{c}{I}_{\text{max}}=\frac{20.0\text{V}}{75.0\Omega }\\ =0.267\text{A}\end{array}$

Conclusion:

Therefore, the maximum current is $0.267\text{A}$.

To determine

 The impedance.

Answer to Problem 33.59AP

 The impedance is $83.9\Omega$.

Explanation of Solution

Given info: The resistance of the circuit is $75.0\Omega$, the frequency is $60.0\text{Hz}$. The voltage drop across the resistor is $20.0\text{V}$, the voltage drop across the inductor is $25.0\text{V}$ and the voltage drop across the capacitor is $15.0\text{V}$.

The expression for the impedance is,

$Z=\frac{R}{\cos \varphi }$

Substitute $75.0\Omega$ for $R$ and $26.6°$ for $\varphi$ in the above expression.

$\begin{array}{c}Z=\frac{75.0\Omega }{\cos \left(26.6°\right)}\\ =83.87\Omega \\ \approx 83.9\Omega \end{array}$

Conclusion:

Therefore, the impedance is $83.9\Omega$.

To determine

 The capacitance.

Answer to Problem 33.59AP

 The capacitance is $47.2\text{μF}$.

Explanation of Solution

Given info: The resistance of the circuit is $75.0\Omega$, the frequency is $60.0\text{Hz}$. The voltage drop across the resistor is $20.0\text{V}$, the voltage drop across the inductor is $25.0\text{V}$ and the voltage drop across the capacitor is $15.0\text{V}$.

The circuit is series $RLC$ circuit, therefore the value of current throughout the circuit is same.

The expression capacitive reactance is,

$X_{\text{C}}=\frac{\Delta {\text{V}}_{\text{C}}}{I_{\text{max}}}$

Substitute $15.0\text{V}$ for $\Delta {\text{V}}_{\text{C}}$ and $0.267\text{A}$ for $I_{\text{max}}$ in the above expression.

$\begin{array}{c}{X}_{\text{C}}=\frac{15.0\text{V}}{0.267\text{A}}\\ =56.18\Omega \end{array}$

The expression capacitive reactance in terms of the capacitance is,

$X_{\text{C}}=\frac{1}{2\pi fC}$

Here,

$f$ is the frequency.

$C$ is the capacitance.

Rearrange the above equation for the value of capacitance.

$C=\frac{1}{2\pi f{X}_{\text{C}}}$

Substitute $56.18\Omega$ for $X_{\text{C}}$ and $60.0\text{Hz}$ for $f$ in the above expression.

$\begin{array}{c}C=\frac{1}{2\pi \left(60.0\text{Hz}\right)\left(56.18\Omega \right)}\\ =47.2\times {10}^{-6}\text{F}\\ =47.2\text{μF}\end{array}$

Conclusion:

Therefore, the capacitance is $47.2\text{μF}$.

To determine

 The inductance of the circuit.

Answer to Problem 33.59AP

 The inductance of the circuit $0.248\text{H}$.

Explanation of Solution

Given info: The resistance of the circuit is $75.0\Omega$, the frequency is $60.0\text{Hz}$. The voltage drop across the resistor is $20.0\text{V}$, the voltage drop across the inductor is $25.0\text{V}$ and the voltage drop across the capacitor is $15.0\text{V}$.

The circuit is series $RLC$ circuit, therefore the value of current throughout the circuit is same.

The expression inductive reactance is,

$X_{\text{L}}=\frac{\Delta {\text{V}}_{\text{L}}}{I_{\text{max}}}$

Substitute $25.0\text{V}$ for $\Delta {\text{V}}_{\text{C}}$ and $0.267\text{A}$ for $I_{\text{max}}$ in the above expression.

$\begin{array}{c}{X}_{\text{L}}=\frac{25.0\text{V}}{0.267\text{A}}\\ =93.63\Omega \end{array}$

Thus the value of inductive reactance is $93.63\Omega$.

The expression inductive reactance in terms of the inductance is,

$X_{\text{L}}=2\pi fL$

Here,

$f$ is the frequency.

$L$ is the capacitance.

Rearrange the above equation for the value of inductance.

$L=\frac{X_{\text{L}}}{2\pi f}$

Substitute $93.63\Omega$ for $X_{\text{C}}$ and $60.0\text{Hz}$ for $f$ in the above expression.

$\begin{array}{c}C=\frac{93.63\Omega }{2\pi \left(60.0\text{Hz}\right)}\\ =0.248\text{H}\\ \approx 0.248\text{H}\end{array}$

Conclusion:

Therefore, the inductance is $0.248\text{H}$.

To determine

 The average power delivered to circuit.

Answer to Problem 33.59AP

 The average power delivered to circuit is $2.67\text{W}$.

Explanation of Solution

Given info: The resistance of the circuit is $75.0\Omega$, the frequency is $60.0\text{Hz}$. The voltage drop across the resistor is $20.0\text{V}$, the voltage drop across the inductor is $25.0\text{V}$ and the voltage drop across the capacitor is $15.0\text{V}$.

The expression for R.M.S value of the current is,

$I_{\text{RMS}}=\frac{I_{\text{max}}}{\sqrt{2}}$

The expression for the average power delivered is,

$P_{\text{avg}}={\left(I_{\text{RMS}}\right)}^{2}R$

Substitute $\frac{I_{\text{max}}}{\sqrt{2}}$ for $I_{\text{RMS}}$ in the above equation.

$\begin{array}{c}{P}_{\text{avg}}={\left(\frac{I_{\text{max}}}{\sqrt{2}}\right)}^{2}R\\ =\frac{{\left(I_{\text{max}}\right)}^2}{2}R\end{array}$

Substitute $0.267\text{A}$ for $I_{\text{max}}$ and $75.0\Omega$ for $R$ in the above equation.

$\begin{array}{c}{P}_{\text{avg}}=\frac{{\left(0.267\text{A}\right)}^2}{2}75.0\Omega \\ =2.67\text{W}\end{array}$

Conclusion:

Therefore, the average power delivered to circuit is $2.67\text{W}$.