
A thermal power plant produces electric power of \[600kW\] at \[4000V\], which is to be transported to a place \[20km\] away from the power plant for consumers' usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with power factor unity. All the currents and voltages mentioned are rms values.
If the direct transmission method with a cable of resistance \[0.4\Omega k{{m}^{-1}}\] is used, the power dissipation (in %) during transmission is?
\[A)20\]
\[B)30\]
\[C)40\]
\[D)50\]
Answer
479.1k+ views
Hint: The power dissipated in a transmission line is proportional to square of the current. Here to find the power, we are given the values of electrical power produced and the voltage. Using the resistance per kilometer, we can find total resistance across \[20km\] cable. By substituting these values in the formula for power dissipated, we can find the percentage loss in power.
Formula used:
\[P={{I}^{2}}R\]
\[I=\dfrac{P}{V}\]
Complete answer:
Here,
Resistance, \[R=0.4\Omega /km\]
Then, total resistance across \[20km\] cable \[=0.4\Omega \times 20=8\Omega \]
We have,
\[P={{I}^{2}}R\] ---------- 1
Where,
\[P\] is the power dissipated
\[I\] is the rms current
\[R\] is the resistance
We have,
Current, \[I=\dfrac{P}{V}\]
Where,
\[P\] is the electrical power
\[V\] is the voltage
Given that, \[P=600kW\] and \[V=4000V\]. Then
Current, \[I=\dfrac{600\times {{10}^{3}}}{4000}=150A\]
Substitute the value of \[I\] and \[R\] in equation 1, we get,
\[P={{150}^{2}}\times 8=180000W=180kW\]
\[Percentage\text{ }loss\text{ }in\text{ }power=\dfrac{180}{600}\times 100=30\%\]
So, the correct answer is “Option B”.
Note:
The transmission over long distances creates power losses. The major part of the energy losses comes from power lines and Joule effect in transformers. The energy is lost as heat in the conductors. And the transmission lines have some amount of resistance. Hence power loss in transmission lines cannot be avoided.
Formula used:
\[P={{I}^{2}}R\]
\[I=\dfrac{P}{V}\]
Complete answer:
Here,
Resistance, \[R=0.4\Omega /km\]
Then, total resistance across \[20km\] cable \[=0.4\Omega \times 20=8\Omega \]
We have,
\[P={{I}^{2}}R\] ---------- 1
Where,
\[P\] is the power dissipated
\[I\] is the rms current
\[R\] is the resistance
We have,
Current, \[I=\dfrac{P}{V}\]
Where,
\[P\] is the electrical power
\[V\] is the voltage
Given that, \[P=600kW\] and \[V=4000V\]. Then
Current, \[I=\dfrac{600\times {{10}^{3}}}{4000}=150A\]
Substitute the value of \[I\] and \[R\] in equation 1, we get,
\[P={{150}^{2}}\times 8=180000W=180kW\]
\[Percentage\text{ }loss\text{ }in\text{ }power=\dfrac{180}{600}\times 100=30\%\]
So, the correct answer is “Option B”.
Note:
The transmission over long distances creates power losses. The major part of the energy losses comes from power lines and Joule effect in transformers. The energy is lost as heat in the conductors. And the transmission lines have some amount of resistance. Hence power loss in transmission lines cannot be avoided.
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