Chapter 21, Problem 32P

(a)

To determine

The impedance of the circuit.

Answer to Problem 32P

The impedance of the circuit is $66.8\text{Ω}$.

Explanation of Solution

Given info: The AC voltage has a form $\Delta v=\left(90.0\text{V}\right)\sin \left(350t\right)$. The resistance is $50.0\text{Ω}$. The capacitance is $25.0\text{μF}$. The inductance is $0.200\text{H}$.

The expression of AC voltage at any instant is given by,

$v=\Delta V_m\sin \left(\omega t\right)$ (1)

• $\Delta V_m$ is the maximum voltage supplied by the AC generator
• $\omega$ is the angular frequency
• $t$ is the time

The given expression of AC voltage is,

$\Delta v=\left(90.0\text{V}\right)\sin \left(350t\right)$ (2)

Comparing (1) and (2), the maximum voltage supplied by the AC source is $90.0\text{V}$. The angular frequency is $350\text{rad}\cdot {\text{s}}^{\text{-1}}$.

Formula to calculate the inductive reactance is,

$X_L=\omega L$

• $\omega$ is the angular frequency of the ac source
• $L$ is the inductance of the inductor

Formula to calculate the capacitive reactance is,

$X_C=\frac{1}{\omega C}$

• $C$ is the capacitance of the capacitor
• $\omega$ is the angular frequency of the ac source

The formula to calculate the impedance of the circuit where the components are in series is given by,

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

• $R$ is the resistance of the resistor

Substituting the expressions for the capacitive and inductive reactance, the impedance of the circuit will be,

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

Substitute $50.0\text{Ω}$ for $R$, $25.0\text{μF}$ for $C$, $2.50\text{H}$ for $L$, $350\text{rad}\cdot {\text{s}}^{\text{-1}}$ for $\omega$ to determine the impedance of the circuit,

$\begin{array}{c}Z=\sqrt{{\left(50.0\text{Ω}\right)}^2+{\left(\left(350\text{rad}\cdot {\text{s}}^{\text{-1}}\right)\left(2.50\text{H}\right)-\frac{1}{\left(350\text{rad}\cdot {\text{s}}^{\text{-1}}\right)\left(25.0\text{μF}\right)}\right)}^2}\\ =66.8\text{Ω}\end{array}$

Conclusion: The impedance of the circuit is $66.8\text{Ω}$.

b)

To determine

The rms current in the circuit.

Answer to Problem 32P

The rms current if the circuit is $0.953\text{A}$.

Explanation of Solution

Given info: The AC voltage has a form $\Delta v=\left(90.0\text{V}\right)\sin \left(350t\right)$. The resistance is $50.0\text{Ω}$. The capacitance is $25.0\text{μF}$. The inductance is $0.200\text{H}$.

Explanation:

The expression of AC voltage at any instant is given by,

$v=\Delta V_m\sin \left(\omega t\right)$ (1)

• $\Delta V_m$ is the maximum voltage supplied by the AC generator
• $\omega$ is the angular frequency
• $t$ is the time

The given expression of AC voltage is,

$\Delta v=\left(90.0\text{V}\right)\sin \left(350t\right)$ (2)

Comparing (1) and (2), the maximum voltage supplied by the AC source is $90.0\text{V}$. The angular frequency is $350\text{rad}\cdot {\text{s}}^{\text{-1}}$.

The rms current in the circuit is given by,

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

• $\Delta V_{rms}$ is the rms voltage of the source
• $Z$ is the impedance

The rms voltage of the source is,

$\Delta V_{rms}=\frac{\Delta V_m}{\sqrt{2}}$

• $\Delta V_m$ is the maximum voltage supplied by the source

Substituting the expression for the rms voltage, the rms current will be,

$I_{rms}=\frac{\Delta V_m}{Z\sqrt{2}}$

Substitute $90.0\text{V}$ for $\Delta V_m$ and $66.8\text{Ω}$ for $Z$ to determine the rms current in the circuit,

$\begin{array}{c}{I}_{rms}=\frac{90.0\text{V}}{\left(66.8\text{Ω}\right)\sqrt{2}}\\ =0.953\text{A}\end{array}$

Conclusion: The rms current if the circuit is $0.953\text{A}$.

c)

To determine

The average power delivered to the circuit.

Answer to Problem 32P

The average power delivered to the circuit is $45.4\text{W}$.

Explanation of Solution

Given info: The AC voltage has a form $\Delta v=\left(90.0\text{V}\right)\sin \left(350t\right)$. The resistance is $50.0\text{Ω}$. The capacitance is $25.0\text{μF}$. The inductance is $0.200\text{H}$.

The average power delivered to a resistor is given by,

$P_a=I_{rms}\Delta V_{rms}\cos \varphi$

• $\varphi$ is the phase angle
• $I_{rms}$ is the rms current flowing in the circuit
• $\Delta V_{rms}$ is the rms voltage

The rms voltage of the source is,

$\Delta V_{rms}=\frac{\Delta V_m}{\sqrt{2}}$

• $\Delta V_m$ is the maximum voltage supplied by the source

Substituting the expression for the rms voltage, the average power can be rewritten as,

$P_a=\frac{I_{rms}\Delta V_m\cos \varphi }{\sqrt{2}}$

Formula to calculate the inductive reactance is,

$X_L=\omega L$

• $\omega$ is the angular frequency of the ac source
• $L$ is the inductance of the inductor

Formula to calculate the capacitive reactance is,

$X_C=\frac{1}{\omega C}$

• $C$ is the capacitance of the capacitor
• $\omega$ is the angular frequency of the ac source

The phase angle of the circuit is given by,

$\varphi ={\tan }^{-1}\left(\frac{X_L-X_C}{R}\right)$

Substituting the expression for inductive reactance and capacitive reactance the expression for the phase angle of the circuit can be rewritten as,

$\varphi ={\tan }^{-1}\left(\frac{\omega L-\frac{1}{\omega C}}{R}\right)$

Substituting the expression for the phase angle, the expression for the average power delivered can be rewritten as,

$P_a=\frac{I_{rms}\Delta V_m}{\sqrt{2}}\cos \left({\tan }^{-1}\left(\frac{\omega L-\frac{1}{\omega C}}{R}\right)\right)$

Substitute $0.953\text{A}$ for $I_{rms}$, $90.0\text{V}$ for $\Delta V_m$, $50.0\text{Ω}$ for $R$, $25.0\text{μF}$ for $C$, $2.50\text{H}$ for $L$, $350\text{rad}\cdot {\text{s}}^{\text{-1}}$ for $\omega$ to determine the average power,

$\begin{array}{c}{P}_a=\frac{\left(0.953\text{A}\right)90.0\text{V}}{\sqrt{2}}\cos \left({\tan }^{-1}\left(\frac{\left(350\text{rad}\cdot {\text{s}}^{\text{-1}}\right)\left(2.50\text{H}\right)-\frac{1}{\left(350\text{rad}\cdot {\text{s}}^{\text{-1}}\right)\left(25.0\text{μF}\right)}}{50.0\text{Ω}}\right)\right)\\ =45.4\text{W}\end{array}$

Conclusion: The average power delivered to the circuit is $45.4\text{W}$.