Chapter 21, Problem 97P

(a)

To determine

The current that would have to flow through the copper wires.

Answer to Problem 97P

The current that would have to flow through the copper wires is $6.7\times {10}^3\text{ A}$ .

Explanation of Solution

Write the equation for the average power.

  $P_{\text{av}}=I_{\text{rms}}{V}_{\text{rms}}$                                                                                                            (I)

Here, $P_{\text{av}}$ is the average power, $I_{\text{rms}}$ is the rms value of the current and $V_{\text{rms}}$ is the rms value of the voltage.

Rewrite the above equation for $I_{\text{rms}}$ .

  $I_{\text{rms}}=\frac{P_{\text{av}}}{V_{\text{rms}}}$                                                                                                            (II)

Conclusion:

Given that the average power is $800\text{ kW}$ and the rms value of the voltage is $120\text{ V}$ .

Substitute $800\text{ kW}$ for $P_{\text{av}}$ and $120\text{ V}$ for $V_{\text{rms}}$ in equation (II) to find $I_{\text{rms}}$ .

  $\begin{array}{c}{I}_{\text{rms}}=\frac{800\text{ kW}\frac{1000\text{ W}}{1\text{ kW}}}{120\text{ V}}\\ =\frac{8.0\times {10}^5\text{ W}}{120\text{ V}}\\ =6667\text{ A}\\ =6.7\times {10}^3\text{ A}\end{array}$

Therefore, the current that would have to flow through the copper wires is $6.7\times {10}^3\text{ A}$ .

(b)

To determine

The power dissipated due to the resistance of the copper wires.

Answer to Problem 97P

All of the power is converted into internal energy in the copper wires.

Explanation of Solution

Write the equation for the power converted in the wires due to the resistance.

  $P_{\text{av}}=I_{\text{rms}}^{2}R$                                                                                                           (III)

Here, $R$ is the resistance of the copper wires.

Conclusion:

Given that the resistance of the copper wires is $12\text{ Ω}$.

Substitute $6667\text{ A}$ for $I_{\text{rms}}$ and $12\text{ Ω}$ for $R$ in equation (III) to find $P_{\text{av}}$ .

  $\begin{array}{c}{P}_{\text{av}}={\left(6667\text{ A}\right)}^2\left(12\text{ Ω}\right)\\ =530000\times {10}^3\text{ W}\\ =530000\text{ kW}\end{array}$

But the average power at which the voltage is sent is $800\text{ kW}$ and $530000\text{ kW}>800\text{ kW}$ . This implies all the power is converted into internal energy due to the resistance of the copper wires.

Thus, all of the power is converted into internal energy in the copper wires since $530000\text{ kW}$ is greater than the power output of the plant.

(c)

To determine

The current flowing through the wires when the transformers are used.

Answer to Problem 97P

The current flowing through the wires when the transformers are used is $17\text{ A}$ .

Explanation of Solution

Equation (II) can be used to determine the value of the current.

Conclusion:

Given that the rms value of the voltage sent is $48\text{ kV}$ .

Substitute $800\text{ kW}$ for $P_{\text{av}}$ and $48\text{ kV}$ for $V_{\text{rms}}$ in equation (II) to find $I_{\text{rms}}$ .

  $\begin{array}{c}{I}_{\text{rms}}=\frac{800\text{ kW}\frac{1000\text{ W}}{1\text{ kW}}}{48\text{ kV}\frac{1000\text{ V}}{1\text{ kV}}}\\ =\frac{8.0\times {10}^5\text{ W}}{48000\text{ V}}\\ =16.7\text{ A}\\ \approx 17\text{ A}\end{array}$

Therefore, the current flowing through the wires when the transformers are used is $17\text{ A}$ .

(d)

To determine

The power dissipated due to the resistance of the wires when the transformers are used and the percent of this value with respect to the total power output of the power plant.

Answer to Problem 97P

The power dissipated due to the resistance of the wires when the transformers are used is $3.3\text{ kW}$ and it is $0.42\text{ \%}$ of the total power output of the power plant.

Explanation of Solution

Equation (III) can be used to determine the power dissipated.

Conclusion:

Substitute $16.7\text{ A}$ for $I_{\text{rms}}$ and $12\text{ Ω}$ for $R$ in equation (III) to find $P_{\text{av}}$ .

  $\begin{array}{c}{P}_{\text{av}}={\left(16.7\text{ A}\right)}^2\left(12\text{ Ω}\right)\\ =3.333\times {10}^3\text{ W}\\ =3.3\text{ kW}\end{array}$

Find the percent of this power loss is of the total power output of the power plant.

$\begin{array}{c}\text{percent}=\frac{3.333\times {10}^3\text{ W}}{800\text{ kW}\frac{1000\text{ W}}{1\text{ kW}}}\times 100\text{ \%}\\ =\frac{3333\text{ W}}{800000\text{ W}}\times 100\text{ \%}\\ =0.42\text{ \%}\end{array}$

Therefore, the power dissipated due to the resistance of the wires when the transformers are used is $3.3\text{ kW}$ and it is $0.42\text{ \%}$ of the total power output of the power plant.

(e)

To determine

The number of secondary turns the transformer should have.

Answer to Problem 97P

The number of secondary turns the transformer should have is $25$ .

Explanation of Solution

Write the equation for a transformer.

  $\frac{{\epsilon }_2}{{\epsilon }_1}=\frac{N_2}{N_1}$

Here, ${\epsilon }_2$ is the voltage produced in the secondary coil, ${\epsilon }_1$ is the voltage produced in the primary coil, $N_2$ is the number of turns in the secondary coil and $N_1$ is the number of turns in the primary coil

Rewrite the above equation for $N_2$ .

  $N_2=\frac{{\epsilon }_2}{{\epsilon }_1}{N}_1$ (IV)

Conclusion:

Given that the number of turns in the primary coil is $10000$ .

Substitute $120\text{ V}$ for ${\epsilon }_2$ , $48\text{ kV}$ for ${\epsilon }_1$ and $10000$ for $N_1$ in equation (IV) to find $N_2$ .

  $\begin{array}{c}{N}_2=\frac{120\text{ V}}{48\text{ kV}\frac{1000\text{ V}}{1\text{ kV}}}\left(10000\right)\\ =\frac{120\text{ V}}{48000\text{ V}}\left(10000\right)\\ =25\end{array}$

Therefore, the number of secondary turns the transformer should have is $25$ .

Therefore, the average power dissipated in the wires when using the transformer is $98\text{ W}$ .