Chapter 29, Problem 41P

(a)

To determine

The RMS current in all branch of the circuit if frequency is 500 Hz.

Answer to Problem 41P

$\boxed{I_{rms}=7.07\text{A}}$ , $\boxed{I_{R,rms}=3.17\text{A}}$ and $\boxed{I_{R_L,rms}=6.31\text{A}}$ .

Explanation of Solution

Given:

Load resistance is $R_L=20\Omega$

Inductance of inductor is $L=3.20\text{mH}$

Resistance of resistor is $R=4\Omega$

Frequency is $f=500\text{Hz}$

Ideal voltage output is $\epsilon =\left(100\text{V}\right)\times \cos \left(2\pi ft\right)$

Formula used:

Phase angle is expressed as

$\delta ={\tan }^{-1}\left(\frac{R_L}{2\pi fL}\right)$…… (1)

Here, $R_L,f,L$ are load resistance, frequency and inductance of inductor.

Resultant reactance for parallel combination is is

$Z=\frac{1}{\sqrt{R^{-2}{}_L+X^{-2}{}_L}}$…… (2)

Here, $X_L$ is inductance of inductor.

Peak current is

$I_{peak}=\frac{{\epsilon }_{peak}}{Z}$…… (3)

Calculation:

Substitute $20\Omega$ for $R_L$ , 500 Hz for $f$ , 3.20 mH for $L$ in equation (1)

$\begin{array}{l}\delta ={\tan }^{-1}\left(\frac{20}{2\pi \times 500\times 3.20\times {10}^{-3}}\right)\\ \delta =63.31°\end{array}$

$\begin{array}{l}{X}_L=2\pi fL\\ X_L=2\times \pi \times 500\times 3.20\times {10}^{-3}\\ X_L=10.05\Omega \end{array}$

Substitute $20\Omega$ for $R_L$ , 10.05 $\Omega$ for $X_L$ in equation (2)

$\begin{array}{l}{Z}_2=\frac{1}{\sqrt{{20}^{-2}+{\left(10.05\right)}^{-2}}}\\ Z_2=8.97\text{A}\end{array}$

Therefore, final value of $Z$ is calculated as

$\begin{array}{l}Z=\sqrt{4^2+{\left(8.97\right)}^2+2\times 4\times \cos \left(63.31\right)}\\ Z=10\Omega \end{array}$

Substitute 100 V for $V_{peak}$ and $10\Omega$ for $Z$ in equation (3)

$\begin{array}{l}{I}_{peak}=\frac{100}{10}\\ I_{peak}=10\text{A}\end{array}$

RMS current is

$\begin{array}{l}{I}_{rms}=\frac{10}{\sqrt{2}}\\ \boxed{I_{rms}=7.07\text{A}}\end{array}$

$\begin{array}{l}{V}_{2,peak}=I_{peak}{Z}_2=10\times 8.97\\ V_{2,peak}=89.7\text{V}\end{array}$

And

$\begin{array}{l}{V}_{2,rms}=\frac{1}{\sqrt{2}}{V}_{2,peak}\\ V_{2,rms}=\frac{1}{\sqrt{2}}\times 89.7\\ V_{2,rms}=63.43\text{V}\end{array}$

The current for resistor and inductor are

$\begin{array}{l}{I}_{R,rms}=\frac{V_{2,rms}}{R_L}\\ I_{R,rms}=\frac{63.43\text{V}}{20}\\ \boxed{I_{R,rms}=3.17\text{A}}\end{array}$

And

$\begin{array}{l}{I}_{R_L,rms}=\frac{V_{2,rms}}{X_L}\\ I_{R_L,rms}=\frac{63.43}{10.05}\\ \boxed{I_{R_L,rms}=6.31\text{A}}\end{array}$

Conclusion:

Hence, the requiredcurrent across each branch are $\boxed{I_{rms}=7.07\text{A}}$ , $\boxed{I_{R,rms}=3.17\text{A}}$ and $\boxed{I_{R_L,rms}=6.31\text{A}}$ .

(b)

To determine

RMS current in all branch of the circuit if frequency is 2000 Hz.

Answer to Problem 41P

The required current across each branch are $\boxed{I_{rms}=3.84\text{A}}$ , $\boxed{I_{R,rms}=3.44\text{A}}$ and $\boxed{I_{R_L,rms}=1.71\text{A}}$ .

Explanation of Solution

Given:

Load resistance is $R_L=20\Omega$

Inductance of inductor is $L=3.20\text{mH}$

Resistance of resistor is $R=4\Omega$

Frequency is $f=2000\text{Hz}$

Ideal voltage output is $\epsilon =\left(100\text{V}\right)\times \cos \left(2\pi ft\right)$

Formula used:

Phase angle is expressed as

$\delta ={\tan }^{-1}\left(\frac{R_L}{2\pi fL}\right)$…… (1)

Here, $R_L,f,L$ are load resistance, frequency and inductance of inductor.

Resultant inductance is

$Z=\frac{1}{\sqrt{R^{-2}{}_L+X^{-2}{}_L}}$…… (2)

Here, $X_L$ is inductance of inductor.

Peak current is

$I_{peak}=\frac{{\epsilon }_{peak}}{Z}$…… (3)

Calculation:

Substitute $20\Omega$ for $R_L$ , 2000 Hz for $f$ , 3.20 $\text{mH}$ for $L$ in equation (1)

$\begin{array}{l}\delta ={\tan }^{-1}\left(\frac{20}{2\pi \times 2000\times 3.20\times {10}^{-3}}\right)\\ \delta =26.4°\end{array}$

$\begin{array}{l}{X}_L=2\pi fL\\ X_L=2\times \pi \times 2000\times 3.20\times {10}^{-3}\\ X_L=40.20\Omega \end{array}$

Substitute $20\Omega$ for $R_L$ , 40.20 $\Omega$ for $X_L$ in equation (2)

$\begin{array}{l}{Z}_2=\frac{1}{\sqrt{{20}^{-2}+{\left(40.20\right)}^{-2}}}\\ Z_2=17.9\Omega \end{array}$

Therefore, final value of $Z$ is calculated as

$\begin{array}{l}Z=\sqrt{4^2+{\left(17.9\right)}^2+2\times 4\times \cos \left(26.4\right)}\\ Z=18.40\Omega \end{array}$

Substitute 100 V for $V_{peak}$ and $18.40\Omega$ for $Z$ in equation (3)

RMS current is

$\begin{array}{l}{I}_{rms}=\frac{5.43}{\sqrt{2}}\\ \boxed{I_{rms}=3.84\text{A}}\end{array}$

$\begin{array}{l}{V}_{2,peak}=I_{peak}{Z}_2=5.43\times 17.90\\ V_{2,peak}=97.19\text{V}\end{array}$

And

$\begin{array}{l}{V}_{2,rms}=\frac{1}{\sqrt{2}}{V}_{2,peak}\\ V_{2,rms}=\frac{1}{\sqrt{2}}\times 97.19\\ V_{2,rms}=68.72\text{V}\end{array}$

The current for resistor and inductor are

$\begin{array}{l}{I}_{R,rms}=\frac{V_{2,rms}}{R_L}\\ I_{R,rms}=\frac{68.72\text{V}}{20}\\ \boxed{I_{R,rms}=3.44\text{A}}\end{array}$

And

$\begin{array}{l}{I}_{R_L,rms}=\frac{V_{2,rms}}{X_L}\\ I_{R_L,rms}=\frac{68.72\text{V}}{40.20}\\ \boxed{I_{R_L,rms}=1.71\text{A}}\end{array}$

Conclusion:

Hence, the required current across each branch are $\boxed{I_{rms}=3.84\text{A}}$ , $\boxed{I_{R,rms}=3.44\text{A}}$ and $\boxed{I_{R_L,rms}=1.71\text{A}}$ .

(c)

To determine

The fraction of the total average power supplied by the ac generator that is delivered to the load resistor if the frequency is 500 Hz.

Answer to Problem 41P

  $\boxed{\frac{P_{R_L}}{P_{R_L}+P_R}=80\%}$

Explanation of Solution

Given:

Load resistance is $R_L=20\Omega$

Inductance of inductor is $L=3.20\text{mH}$

Resistance of resistor is $R=4\Omega$

Frequency is $f=500\text{Hz}$

RMS current is $I_{rms}=7.07\text{A}$

RMS current across $R_L$ is $I_{R_L}=6.31\text{A}$

Ideal voltage output is $\epsilon =\left(100\text{V}\right)\times \cos \left(2\pi ft\right)$

Formula used:

The fraction of the total power delivered by the source that is dissipated in load resistor can be expressed as

  $\frac{P_{R_L}}{P_{R_L}+P_R}=\left(1+\frac{P_R}{P_R{}_{{}_L}}\right)$

$\frac{P_{R_L}}{P_{R_L}+P_R}={\left(1+\frac{I^2{}_{rms}}{I^2{}_{R_L,rms}{R}_L}\right)}^{-1}$…… (1)

Here, $P_{R_L,}{I}_{rms},P_R,R_L$ and $R$ are power dissipated across load resistor, RMS current, power dissipated across resistor, resistance across inductor and resistance of resistor respectively.

Calculation:

Substitute 500 Hz for $f$ , $4\Omega$ for $R$ , $20\Omega$ for $R_L$ , $7.07\text{A}$ for $I_{rms}$ and $6.31\text{A}$ for $I_{R_L}$ in equation (1)

  $\begin{array}{l}\frac{P_{R_L}}{P_{R_L}+P_R}={\left(1+\frac{{\left(7.07\right)}^2\times 4}{{\left(6.31\right)}^2\times 20}\right)}^{-1}\\ \frac{P_{R_L}}{P_{R_L}+P_R}=0.80\\ \boxed{\frac{P_{R_L}}{P_{R_L}+P_R}=80\%}\end{array}$

Conclusion:

Hence, the required the fraction of the total average power supplied by the ac generator that is delivered to the load resistor if the frequency is 500 Hz is $\boxed{\frac{P_{R_L}}{P_{R_L}+P_R}=80\%}$ .

(d)

To determine

The fraction of the total average power supplied by the ac generator that is delivered to the load resistor if the frequency is 2000 Hz.

Answer to Problem 41P

  $\boxed{\frac{P_{R_L}}{P_{R_L}+P_R}=50\%}$

Explanation of Solution

Given:

Load resistance is $R_L=20\Omega$

Inductance of inductor is $L=3.20\text{mH}$

Resistance of resistor is $R=4\Omega$

Frequency is $f=2000\text{Hz}$

RMS current is $I_{rms}=3.84\text{A}$

RMS current across $R_L$ is $I_{R_L}=1.71\text{A}$

Ideal voltage output is $\epsilon =\left(100\text{V}\right)\times \cos \left(2\pi ft\right)$

Formula used:

The fraction of the total power delivered by the source that is dissipated in load resistor can be expressed as

  $\frac{P_{R_L}}{P_{R_L}+P_R}=\left(1+\frac{P_R}{P_R{}_{{}_L}}\right)$

$\frac{P_{R_L}}{P_{R_L}+P_R}={\left(1+\frac{I^2{}_{rms}}{I^2{}_{R_L,rms}{R}_L}\right)}^{-1}$…… (1)

Here, $P_{R_L,}{I}_{rms},P_R,R_L$ and $R$ are power dissipated across load resistor, RMS current, power dissipated across resistor, resistance across inductor and resistance of resistor respectively.

Calculation:

Substitute 2000 Hz for $f$ , $4\Omega$ for $R$ , $20\Omega$ for $R_L$ , $3.84\text{A}$ for $I_{rms}$ and $1.71\text{A}$ for $I_{R_L}$ in equation (1)

  $\begin{array}{l}\frac{P_{R_L}}{P_{R_L}+P_R}={\left(1+\frac{{\left(3.84\right)}^2\times 4}{{\left(1.71\right)}^2\times 20}\right)}^{-1}\\ \frac{P_{R_L}}{P_{R_L}+P_R}=0.50\\ \boxed{\frac{P_{R_L}}{P_{R_L}+P_R}=50\%}\end{array}$

Conclusion:

Hence, the required the fraction of the total average power supplied by the ac generator that is delivered to the load resistor if the frequency is 2000 Hz is $\boxed{\frac{P_{R_L}}{P_{R_L}+P_R}=50\%}$ .