Chapter 32, Problem 16P

An AC source with ΔV_max = 150 V and f = 50.0 Hz is connected between points a and d in Figure P32.16. Calculate the maximum voltages between (a) points a and b, (b) points b and c, (c) points c and d, and (d) points b and d.

Figure P32.16 Problems 16 and 51.

To determine

The maximum voltage between points $a$ and $b$.

Answer to Problem 16P

The maximum voltage between points $a$ and $b$ is $146.24\text{V}$.

Explanation of Solution

Given Information: The value of resistance is $40.0\Omega$, the value of capacitance is $65.0\text{μF}$, the value of inductor is $185\text{mH}$, the frequency of an AC source is $50.0\text{Hz}$, maximum voltage across the AC source is $150\text{V}$.

Formula to calculate the value of angular velocity is,

$\omega =2\pi f$

Here,

$\omega$ is the value of angular velocity.

$f$ is the frequency that gives this current.

Formula to calculate the value of capacitive reactance is,

$X_{\text{C}}=\frac{1}{\omega C}$ (1)

Here,

$X_{\text{C}}$ is the value of capacitive reactance.

$C$ is the value of capacitance.

Substitute $2\pi f$ for $\omega$ in equation (1) to find $X_{\text{C}}$,

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

Substitute $65.0\text{μF}$ for $C$, $50.0\text{Hz}$ for $f$ in equation (2) to find $X_{\text{C}}$,

$\begin{array}{c}{X}_{\text{C}}=\frac{1}{2\pi \times 50.0\text{Hz}\times \left(65.0\text{μF}\times \frac{1\text{F}}{1000000\text{μF}}\right)}\\ =48.97\text{Ω}\end{array}$

Thus, the value of capacitive reactance is $48.97\text{Ω}$.

Formula to calculate the value of inductive reactance is,

$X_L=\omega L$ (3)

Here,

$X_L$ is the value of inductive reactance.

$L$ is the value of inductor.

Substitute $2\pi f$ for $\omega$ in equation (3) to find $X_L$,

$X_L=2\pi fL$ (4)

Substitute $185\text{mH}$ for $L$, $50.0\text{Hz}$ for $f$ in equation (4) to find $X_L$,

$\begin{array}{c}{X}_L=2\pi \times 50.0\text{Hz}\times \left(185\text{mH}\times \frac{1\text{H}}{1000\text{mH}}\right)\\ =58.119\text{Ω}\\ \approx 58.12\text{Ω}\end{array}$

Thus, the value of inductive reactance is $58.12\text{Ω}$.

In $L,C,R$ circuit, $E$ leads $I$ by $\varphi$. The net resistance offered by the $L,C,R$ in the path of AC is called Impedence of the circuit.

Formula to calculate the value of impedence of the circuit is,

$Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$ (5)

Here,

$Z$ is the impedence of the circuit.

$R$ is the value of resistance.

Substitute $58.12\text{Ω}$ for $X_L$, $48.97\text{Ω}$ for $X_{\text{C}}$, $40.0\Omega$ for $R$ in equation (5) to find $Z$,

$\begin{array}{c}Z=\sqrt{{\left(40.0\Omega \right)}^2+{\left(58.12\text{Ω}-48.97\text{Ω}\right)}^2}\\ =41.03\text{Ω}\end{array}$

Thus, the value of impedence of the circuit is $41.03\text{Ω}$.

Formula to calculate the maximum current across the $L,C,R$ circuit is,

$I_{\text{max}}=\frac{\Delta V_{\text{max}}}{Z}$ (6)

Here,

$I_{\text{max}}$ is the maximum current across the $L,C,R$ circuit.

$\Delta V_{\text{max}}$ is the maximum voltage across the AC source.

Substitute $150\text{V}$ for $\Delta V_{\text{max}}$, $41.03\text{Ω}$ for $Z$ in equation (6) to find $I_{\text{max}}$,

$\begin{array}{c}{I}_{\text{max}}=\frac{150\text{V}}{41.03\text{Ω}}\\ =3.6555\text{A}\\ \approx 3.656\text{A}\end{array}$

Thus, the maximum current across the $L,C,R$ circuit is $3.656\text{A}$.

Formula to calculate the maximum voltage between points $a$ and $b$ is,

$V_{\text{ab}}=I_{\text{max}}R$ (7)

Here,

$V_{\text{ab}}$ is the maximum voltage between points $a$ and $b$.

Substitute $3.656\text{A}$ for $I_{\text{max}}$, $40.0\Omega$ for $R$ in equation (7) to find $V_{\text{ab}}$,

$\begin{array}{c}{V}_{\text{ab}}=3.656\text{A}\times 40.0\Omega \\ =146.24\text{V}\end{array}$

Thus, the maximum voltage between points $a$ and $b$ is $146.24\text{V}$.

Conclusion:

Therefore, the maximum voltage between points $a$ and $b$ is $146.24\text{V}$.

To determine

The maximum voltage between points $b$ and $c$.

Answer to Problem 16P

The maximum voltage between points $b$ and $c$ is $212.49\text{V}$.

Explanation of Solution

Given Information: The value of resistance is $40.0\Omega$, the value of capacitance is $65.0\text{μF}$, the value of inductor is $185\text{mH}$, the frequency of an AC source is $50.0\text{Hz}$, maximum voltage across the AC source is $150\text{V}$.

Formula to calculate the maximum voltage between points $b$ and $c$ is,

$V_{\text{bc}}=I_{\text{max}}{X}_L$ (8)

Here,

$V_{\text{bc}}$ is the maximum voltage between points $b$ and $c$.

Substitute $3.656\text{A}$ for $I_{\text{max}}$, $58.12\text{Ω}$ for $X_L$ in equation (8) to find $V_{\text{bc}}$,

$\begin{array}{c}{V}_{\text{bc}}=3.656\text{A}\times 58.12\text{Ω}\\ =212.486\text{V}\\ \approx 212.49\text{V}\end{array}$

Thus, the maximum voltage between points $b$ and $c$ is $212.49\text{V}$.

Conclusion:

Therefore, the maximum voltage between points $b$ and $c$ is $212.49\text{V}$.

To determine

The maximum voltage between points $c$ and $d$.

Answer to Problem 16P

The maximum voltage between points $c$ and $d$ is $179.03\text{V}$.

Explanation of Solution

Given Information: The value of resistance is $40.0\Omega$, the value of capacitance is $65.0\text{μF}$, the value of inductor is $185\text{mH}$, the frequency of an AC source is $50.0\text{Hz}$, maximum voltage across the AC source is $150\text{V}$.

Formula to calculate the maximum voltage between points $c$ and $d$ is,

$V_{\text{cd}}=I_{\text{max}}{X}_{\text{C}}$ (9)

Here,

$V_{\text{cd}}$ is the maximum voltage between points $c$ and $d$.

Substitute $3.656\text{A}$ for $I_{\text{max}}$, $48.97\text{Ω}$ for $X_{\text{C}}$ in equation (9) to find $V_{\text{cd}}$,

$\begin{array}{c}{V}_{\text{cd}}=3.656\text{A}\times 48.97\text{Ω}\\ =179.03\text{V}\end{array}$

Thus, the maximum voltage between points $c$ and $d$ is $179.03\text{V}$.

Conclusion:

Therefore, the maximum voltage between points $c$ and $d$ is $179.03\text{V}$.

To determine

The maximum voltage between points $b$ and $d$.

Answer to Problem 16P

The maximum voltage between points $b$ and $d$ is $33.45\text{V}$.

Explanation of Solution

Given Information: The value of resistance is $40.0\Omega$, the value of capacitance is $65.0\text{μF}$, the value of inductor is $185\text{mH}$, the frequency of an AC source is $50.0\text{Hz}$, maximum voltage across the AC source is $150\text{V}$.

Formula to calculate the maximum voltage between points $b$ and $d$ is,

$V_{\text{bd}}=I_{\text{max}}\left(X_L-X_C\right)$ (10)

Here,

$V_{\text{bd}}$ is the maximum voltage between points $b$ and $d$.

Substitute $3.656\text{A}$ for $I_{\text{max}}$, $58.12\text{Ω}$ for $X_L$, $48.97\text{Ω}$ for $X_{\text{C}}$ in equation (10) to find $V_{\text{bd}}$,

$\begin{array}{c}{V}_{\text{bd}}=3.656\text{A}\times \left(58.12\text{Ω}-48.97\text{Ω}\right)\\ =33.45\text{V}\end{array}$

Thus, the maximum voltage between points $b$ and $d$ is $33.45\text{V}$.

Conclusion:

Therefore, the maximum voltage between points $b$ and $d$ is $33.45\text{V}$.