Chapter 29, Problem 42P

(a)

To determine

The average power delivered to each resistor if the driving frequency is 100 Hz.

Answer to Problem 42P

The required average powers are $\boxed{P_1=45.6\text{W}}$ and $\boxed{P_2=55.6\text{W}}$ .

Explanation of Solution

Given:

Ideal AC voltage source is ${\epsilon }_1=\left(20\text{V}\right)\cos \left(2\pi ft\right)$

An ideal battery (whose EMF is ${\epsilon }_2$ ) is ${\epsilon }_2=16\text{V}$

Resistances of two resistor are $R_1=10\Omega$ and $R_2=8\Omega$

Inductance of inductor is $L=6\text{mH}$

Driving frequency is $f=100\text{Hz}$

Formula used:

Total power delivered to $R_1$ and $R_2$ is expressed as

$P_1=P_{1,dc}+P_{1,ac}$…… (1)

And

$P_2=P_{2,dc}+P_{2,ac}$…… (2)

Here, $P_{dc}$ and $P_{dc}$ are dc and ac power respectively

Expression for power is

$P=\frac{{\epsilon }^2}{R}$…… (3)

Here, $R$ and $\epsilon$ are resistance and voltage.

Calculation:

Dc powers delivered to the resistor whose resistances are $R_1\text{and}{R}_2$ are

$P_{1,dc}=\frac{{\epsilon }_2{}^2}{R_1}$….. (4)

And

$P_{2,dc}=\frac{{\epsilon }_2{}^2}{R_2}$…… (5)

Average AC power is

$P_{1,dc}=\frac{{\epsilon }^2{}_{1,rms}}{R_1}=\frac{{\epsilon }^2{}_{1,peak}}{2R_1}$

$P_{1,dc}=\frac{{\epsilon }^2{}_{1,peak}}{2R_1}$…… (6)

Apply Kirchhoff’s loop law,

$R_1{I}_1-Z_2{I}_2=0$…… (7)

Solve equation (7)

$\begin{array}{l}{I}_2=\frac{R_1}{Z_2}{I}_1\\ I_2=\frac{R_1{\epsilon }_{1,peak}}{Z_2{R}_1}\end{array}$

$I_2=\frac{{\epsilon }_{1,peak}}{Z_2}$…… (8)

Average ac power delivered is calculated as

$\begin{array}{l}{P}_{2,ac}=\frac{1}{2}{I}^2{}_{2,rms}{R}_2\\ P_{2,ac}=\frac{1}{2}{\left(\frac{{\epsilon }_{1,peak}}{Z_2}\right)}^2{R}_2\end{array}$

$P_{2,ac}=\frac{{\epsilon }^2{}_{1,peak}{R}_2}{2Z^2{}_2}$…… (9)

Substitute $\frac{{\epsilon }_2{}^2}{R_1}$ for $P_{1,dc}$ , $\frac{{\epsilon }_2{}^2}{R_2}$ for $P_{2,dc}$ , $\frac{{\epsilon }^2{}_{1,peak}}{2R_1}$ for $P_{1,dc}$ , $\frac{{\epsilon }^2{}_{1,peak}{R}_2}{2Z^2{}_2}$ for $P_{2,ac}$ in equation (1) and (2)

$P_1=\frac{{\epsilon }_2{}^2}{R_1}+\frac{{\epsilon }^2{}_{1,peak}}{2R_1}$…… (10)

And

$P_2=\frac{{\epsilon }_2{}^2}{R_2}+\frac{{\epsilon }^2{}_{1,peak}{R}_2}{2Z^2{}_2}$…… (11)

Substitute 16 V for ${\epsilon }_2$ ,20 V for ${\epsilon }_{1,peak}$ , $10\Omega$ for $R_1$ , $8\Omega$ for $R_2$ , 100 Hz for $f$ and $6\text{mH}$ for $L$ ;in equation (10) and (11)

$\begin{array}{l}{P}_1=\frac{{16}^2}{10}+\frac{{20}^2}{2\times 10}\\ \boxed{P_1=45.6\text{W}}\end{array}$

And

$\begin{array}{l}{P}_2=\frac{{16}^2}{8}+\frac{{20}^2\times 8}{2\times \left[8^2+2\times \pi \times 100\times 6\times {10}^{-3}\right]}\\ \boxed{P_2=55.6\text{W}}\end{array}$

Conclusion:

Hence, the requiredaverage powers are $\boxed{P_1=45.6\text{W}}$ and $\boxed{P_2=55.6\text{W}}$ .

(b)

To determine

The average power delivered to each resistor if the driving frequency is 200 Hz.

Answer to Problem 42P

The required average power are $\boxed{P_1=45.6\text{W}}$ and $\boxed{P_2=54.37\text{W}}$ .

Explanation of Solution

Given:

Ideal AC voltage source is ${\epsilon }_1=\left(20\text{V}\right)\cos \left(2\pi ft\right)$

An ideal battery (whose EMF is ${\epsilon }_2$ ) is ${\epsilon }_2=16\text{V}$

Resistances of two resistor are $R_1=10\Omega$ and $R_2=8\Omega$

Inductance of inductor is $L=6\text{mH}$

Driving frequency is $f=200\text{Hz}$

Formula used:

Total power delivered to $R_1$ and $R_2$ is expressed as

$P_1=P_{1,dc}+P_{1,ac}$…… (1)

And

$P_2=P_{2,dc}+P_{2,ac}$…… (2)

Here, $P_{dc}$ and $P_{dc}$ are dc and ac power respectively

Expression for power is

$P=\frac{{\epsilon }^2}{R}$…… (3)

Here, $R$ and $\epsilon$ are resistance and voltage.

Calculation:

Dc powers delivered to the resistor whose resistances are $R_1\text{and}{R}_2$ are

$P_{1,dc}=\frac{{\epsilon }_2{}^2}{R_1}$….. (4)

And

$P_{2,dc}=\frac{{\epsilon }_2{}^2}{R_2}$…… (5)

Average AC power is

$P_{1,dc}=\frac{{\epsilon }^2{}_{1,rms}}{R_1}=\frac{{\epsilon }^2{}_{1,peak}}{2R_1}$

$P_{1,dc}=\frac{{\epsilon }^2{}_{1,peak}}{2R_1}$…… (6)

Apply Kirchhoff’s loop law,

$R_1{I}_1-Z_2{I}_2=0$…… (7)

Solve equation (7)

$\begin{array}{l}{I}_2=\frac{R_1}{Z_2}{I}_1\\ I_2=\frac{R_1{\epsilon }_{1,peak}}{Z_2{R}_1}\end{array}$

$I_2=\frac{{\epsilon }_{1,peak}}{Z_2}$…… (8)

Average ac power delivered is calculated as

$\begin{array}{l}{P}_{2,ac}=\frac{1}{2}{I}^2{}_{2,rms}{R}_2\\ P_{2,ac}=\frac{1}{2}{\left(\frac{{\epsilon }_{1,peak}}{Z_2}\right)}^2{R}_2\end{array}$

$P_{2,ac}=\frac{{\epsilon }^2{}_{1,peak}{R}_2}{2Z^2{}_2}$…… (9)

Substitute $\frac{{\epsilon }_2{}^2}{R_1}$ for $P_{1,dc}$ , $\frac{{\epsilon }_2{}^2}{R_2}$ for $P_{2,dc}$ , $\frac{{\epsilon }^2{}_{1,peak}}{2R_1}$ for $P_{1,dc}$ , $\frac{{\epsilon }^2{}_{1,peak}{R}_2}{2Z^2{}_2}$ for $P_{2,ac}$ in equation (1) and (2)

$P_1=\frac{{\epsilon }_2{}^2}{R_1}+\frac{{\epsilon }^2{}_{1,peak}}{2R_1}$…… (10)

And

$P_2=\frac{{\epsilon }_2{}^2}{R_2}+\frac{{\epsilon }^2{}_{1,peak}{R}_2}{2Z^2{}_2}$…… (11)

Substitute 16 V for ${\epsilon }_2$ , 20 V for ${\epsilon }_{1,peak}$ , $10\Omega$ for $R_1$ , $8\Omega$ for $R_2$ , 200 Hz for $f$ and $6\text{mH}$ for $L$ in equation (10) and (11)

$\begin{array}{l}{P}_1=\frac{{16}^2}{10}+\frac{{20}^2}{2\times 10}\\ \boxed{P_1=45.6\text{W}}\end{array}$

And

$\begin{array}{l}{P}_2=\frac{{16}^2}{8}+\frac{{20}^2\times 8}{2\times \left[8^2+2\times \pi \times 200\times 6\times {10}^{-3}\right]}\\ \boxed{P_2=54.37\text{W}}\end{array}$

Conclusion:

Hence, the required average powers are $\boxed{P_1=45.6\text{W}}$ and $\boxed{P_2=54.37\text{W}}$ .

(c)

To determine

The average power delivered to each resistor if the driving frequency is 800 Hz.

Answer to Problem 42P

The required average powerare $\boxed{P_1=45.6\text{W}}$ and $\boxed{P_2=48.99\text{W}}$ .

Explanation of Solution

Given:

Ideal AC voltage source is ${\epsilon }_1=\left(20\text{V}\right)\cos \left(2\pi ft\right)$

An ideal battery (whose EMF is ${\epsilon }_2$ ) is ${\epsilon }_2=16\text{V}$

Resistances of two resistor are $R_1=10\Omega$ and $R_2=8\Omega$

Inductance of inductor is $L=6\text{mH}$

Driving frequency is $f=800\text{Hz}$

Formula used:

Total power delivered to $R_1$ and $R_2$ is expressed as

$P_1=P_{1,dc}+P_{1,ac}$…… (1)

And

$P_2=P_{2,dc}+P_{2,ac}$…… (2)

Here, $P_{dc}$ and $P_{dc}$ are dc and ac power respectively

Expression for power is

$P=\frac{{\epsilon }^2}{R}$…… (3)

Here, $R$ and $\epsilon$ are resistance and voltage.

Calculation:

Dc powers delivered to the resistor whose resistances are $R_1\text{and}{R}_2$ are

$P_{1,dc}=\frac{{\epsilon }_2{}^2}{R_1}$….. (4)

And

$P_{2,dc}=\frac{{\epsilon }_2{}^2}{R_2}$…… (5)

Average AC power is

$P_{1,dc}=\frac{{\epsilon }^2{}_{1,rms}}{R_1}=\frac{{\epsilon }^2{}_{1,peak}}{2R_1}$

$P_{1,dc}=\frac{{\epsilon }^2{}_{1,peak}}{2R_1}$…… (6)

Apply Kirchhoff’s loop law,

$R_1{I}_1-Z_2{I}_2=0$…… (7)

Solve equation (7)

$\begin{array}{l}{I}_2=\frac{R_1}{Z_2}{I}_1\\ I_2=\frac{R_1{\epsilon }_{1,peak}}{Z_2{R}_1}\end{array}$

$I_2=\frac{{\epsilon }_{1,peak}}{Z_2}$…… (8)

Average ac power delivered is calculated as

$\begin{array}{l}{P}_{2,ac}=\frac{1}{2}{I}^2{}_{2,rms}{R}_2\\ P_{2,ac}=\frac{1}{2}{\left(\frac{{\epsilon }_{1,peak}}{Z_2}\right)}^2{R}_2\end{array}$

$P_{2,ac}=\frac{{\epsilon }^2{}_{1,peak}{R}_2}{2Z^2{}_2}$…… (9)

Substitute $\frac{{\epsilon }_2{}^2}{R_1}$ for $P_{1,dc}$ , $\frac{{\epsilon }_2{}^2}{R_2}$ for $P_{2,dc}$ , $\frac{{\epsilon }^2{}_{1,peak}}{2R_1}$ for $P_{1,dc}$ , $\frac{{\epsilon }^2{}_{1,peak}{R}_2}{2Z^2{}_2}$ for $P_{2,ac}$ in equation (1) and (2)

$P_1=\frac{{\epsilon }_2{}^2}{R_1}+\frac{{\epsilon }^2{}_{1,peak}}{2R_1}$…… (10)

And

$P_2=\frac{{\epsilon }_2{}^2}{R_2}+\frac{{\epsilon }^2{}_{1,peak}{R}_2}{2Z^2{}_2}$…… (11)

Substitute 16 V for ${\epsilon }_2$ , 20 V for ${\epsilon }_{1,peak}$ , $10\Omega$ for $R_1$ , $8\Omega$ for $R_2$ , 800 Hz for $f$ and $6\text{mH}$ for $L$ in equation (10) and (11)

$\begin{array}{l}{P}_1=\frac{{16}^2}{10}+\frac{{20}^2}{2\times 10}\\ \boxed{P_1=45.6\text{W}}\end{array}$

And

$\begin{array}{l}{P}_2=\frac{{16}^2}{8}+\frac{{20}^2\times 8}{2\times \left[8^2+2\times \pi \times 800\times 6\times {10}^{-3}\right]}\\ \boxed{P_2=48.99\text{W}}\end{array}$

Conclusion:

Hence, the required average powers are $\boxed{P_1=45.6\text{W}}$ and $\boxed{P_2=48.99\text{W}}$ .