Figure P 22.71 shows a power station (represented by the AC...

Figure $\mathrm{P} 22.71$ shows a power station (represented by the AC source) connected to a power line $R_{\text {line }}$ and a city represented by $R_{\text {city. }}$ Suppose the power line has a resistance of $0.30 \Omega / \mathrm{km}$ and is 500 km long, and the current in the line is 1000 A. a. If the voltage at the power station is $800 \mathrm{kV}$ (rms), what is $R_{\text {clyy }}$ ? Also find the voltage at the city, the power delivered to the city, and $P_{\lim _{0}} / P_{\text {cily }}$ (the ratio of the power dissipated in the line to the power dissipated in the city). b. Repeat part (a), but now assume the voltage at the power station is $400 \mathrm{kV}$ (rms). Your answer for $P_{\text {line }} / P_{\text {aly } y}$ will illustrate again why it is preferable to use high voltages to transmit electrical power over long distances.

Figure $\mathrm{P} 22.71$ shows a power station (represented by the AC source) connected to a power line $R_{\text {line }}$ and a city represented by $R_{\text {city. }}$ Suppose the power line has a resistance of $0.30 \Omega / \mathrm{km}$ and is 500 km long, and the current in the line is 1000 A.
a. If the voltage at the power station is $800 \mathrm{kV}$ (rms), what is $R_{\text {clyy }}$ ? Also find the voltage at the city, the power delivered to the city, and $P_{\lim _{0}} / P_{\text {cily }}$ (the ratio of the power dissipated in the line
to the power dissipated in the city).
b. Repeat part (a), but now assume the voltage at the power station is $400 \mathrm{kV}$ (rms). Your answer for $P_{\text {line }} / P_{\text {aly } y}$ will illustrate again why it is preferable to use high voltages to transmit electrical power over long distances.

Key Concepts

High Voltage Transmission

High voltage transmission is the practice of increasing the voltage to lower the current for a fixed power level, which in turn minimizes resistive losses (since losses scale with the square of the current). This method is preferred for long-distance power transmission as it greatly enhances efficiency, reducing the energy wasted as heat in transmission lines.

Resistive Power Loss

This concept refers to the energy lost as heat in a conductor due to its resistance, which is quantified by the formula P_loss = I²R. In power transmission lines, resistive losses play a critical role because even small resistances can result in significant energy loss when large currents are used over long distances.

Voltage Drop

In any circuit, especially in long transmission lines, the voltage drop is the reduction in voltage that occurs as current flows through the resistance of the conductor. This drop, calculated as ?V = I × R_line, directly impacts the voltage available at the load (in this case, the city) and is a key factor in assessing transmission efficiency.

RMS Voltage in AC Circuits

Root Mean Square (RMS) voltage is a crucial concept in AC power systems because it provides a measure of the effective voltage. It allows for the comparison and calculation of power in AC circuits in a manner analogous to direct current (DC) systems, and is essential for determining the performance and efficiency of power transmission systems.

Power Transmission Efficiency

This concept deals with the ratio of the power delivered to the load to the total power supplied by the source. It takes into account losses in the transmission line, mainly due to resistive heating. High efficiency in power transmission is critical for minimizing energy losses over long distances, which is a primary concern in electrical power engineering.

Transcript

00:01 Hello everyone, let us do the following question.

00:03 In the question we have given a power station connected to a power line which is having a resistance are suppose the power line as a resistance of power line is given as 0 .30 oom per kilometer.

00:22 The distance is given as 500 kilometers.

00:26 Distance is given as 500 km and here current is given current is given 1000 ampere if the voltage at the power station is 800 kovol we have to find the value of r and also find the voltage at the city we have to find the given terms if we'll find this value if we'll find this value now we will find power we'll solve this question we know that firstly we will find the resistance of the line firstly, we will find the resistance of the line.

01:05 Firstly, we will find the resistance of the line.

01:12 Here we will say that resistance of line is given as 0 .3 oom kilometer, 0 .3 om kilometer into 500 kilometers.

01:26 Is given as 150 om and if we have to find the equivalent resistance if we have the find the equivalent resistance this is equals to 150 om plus our resistance of city resistance of city circuit next if we have to find the resistance for circuit this is v line divided by i line this is equals to 150 plus r into city.

01:59 Here if we have to find the resistance of our city, this is 8 00 volt divided by 1000 ampere.

02:11 This is minus 150 oom.

02:14 After solving this we will get 650 oom.

02:18 Now we find the voltage at the city by setting the two current equations by setting by setting to current equations we will get v line by r circuit is equal to bct by ict.

02:38 It will solve this equation.

02:41 Here we will get value for cct here we will get value for cct is equal to v line into rct divided by r circuit.

02:53 It will put all the value we will get 650 kilo volt because we will say current is same throughout next the ratio of electric power dissipated we have to find the ratio of power dissipated we have to find the ratio of power dissipated is essentially the ratio of resistance is essentially the ratio of resistance if we have to find the ratio of resistance, we will find here power of line divided by power of city.

03:33 This is i square r into line.

03:37 In the denominator we will line i square city.

03:41 I square will be cancelled with i square.

03:43 Here we will get i line divided by ir city.

03:48 It will get 150 oom divided by 650 o 'em.

03:53 This is 0 .23...

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