Question

\[1{\rm{ MW}}\] power is to be delivered from a power station to a town \[10{\rm{ km}}\] away. One uses a pair of Cu wires of radius \[0.5{\rm{ cm}}\] for this purpose. Calculate the fraction of ohmic losses to power transmitted if a step-up transformer is used to boost the voltage to \[11000{\rm{ V}}\], power transmitted, then a step-down transformer is used to bring voltage to \[220{\rm{ V}}\]. (\[{\rho _{Cu}} = 1.7 \times {10^{ - 8}}{\rm{ }}\Omega {\rm{m}}\]).

Answer 1

Hint: Resistance of the copper wire is directly proportional to its length and inversely proportional to the cross-sectional area. We will calculate the resistance so that it can be substituted in the expression for power loss which is the product of resistance and square of the current flowing through the wire. We know that the fraction of power loss is equal to the ratio of power loss to its initial value.

Complete step by step answer:
Given:
The power to be delivered to a town is \[P = 1{\rm{ MW}} = {10^6}{\rm{ W}}\].
The distance of the town from the power station is \[l = 10{\rm{ km}} = 10{\rm{ km}} \times \left( {\dfrac{{1000{\rm{ m}}}}{{{\rm{km}}}}} \right) = 10000{\rm{ m}}\].
The radius of the copper wire used for power transmission is \[r = 0.5{\rm{ cm}} \times \left( {\dfrac{{\rm{m}}}{{100{\rm{ cm}}}}} \right) = 0.5 \times {10^{ - 2}}{\rm{ m}}\].
The value of voltage achieved using a step-down transformer is \[{V_1} = 220{\rm{ V}}\].
The value of voltage achieved using a step-up transformer is \[{V_2} = 11000{\rm{ V}}\].
The resistivity of copper wire is \[{\rho _{Cu}} = 1.7 \times {10^{ - 8}}{\rm{ }}\Omega {\rm{m}}\].
We are required to find the fraction of losses of power transmitted.
We can easily understand that the length of copper wire to be used will be equal to the distance between town and power station. But it is given that a pair of copper wire is used to transmit power; therefore, we can say that the total length of the wire will be twice of its initial value.
\[L = 2l\]
On substituting \[10000{\rm{ m}}\] for l in the above expression, we get:
\[
L = 2\left( {10000{\rm{ m}}} \right)\\
\Rightarrow L = 20000{\rm{ m}}
\]
Let us write the expression for resistance present in copper wire.
\[R = \dfrac{{{\rho _{Cu}}L}}{A}\]……(1)
Here A is the area of cross-section of copper wire, and it is given as:
\[A = \pi {r^2}\]
On substituting \[0.5 \times {10^{ - 2}}{\rm{ m}}\] for r in the above expression, we get:
\[
A = \pi {\left( {0.5 \times {{10}^{ - 2}}{\rm{ m}}} \right)^2}\\
\Rightarrow A = 7.85 \times {10^{ - 5}}{\rm{ }}{{\rm{m}}^2}
\]
On substituting \[1.7 \times {10^{ - 8}}{\rm{ }}\Omega {\rm{m}}\] for \[{\rho _{Cu}}\], \[20000{\rm{ m}}\] for L and \[7.85 \times {10^{ - 5}}{\rm{ }}{{\rm{m}}^2}\] for A in equation (1), we get:
\[
R = \dfrac{{\left( {1.7 \times {{10}^{ - 8}}{\rm{ }}\Omega {\rm{m}}} \right)\left( {20000{\rm{ m}}} \right)}}{{\left( {7.85 \times {{10}^{ - 5}}{\rm{ }}{{\rm{m}}^2}} \right)}}\\
\Rightarrow R = 4.33{\rm{ }}\Omega
\]
Let us write the expression for current when the voltage is stepped down using a step-down transformer.
\[{I_1} = \dfrac{P}{{{V_1}}}\]
Here \[{I_1}\] is the current during the step-down stage.
On substituting \[{10^6}{\rm{ W}}\] for P and \[220{\rm{ V}}\] for \[{V_1}\] in the above expression, we get:
\[
{I_1} = \dfrac{{{{10}^6}{\rm{ W}}}}{{220{\rm{ V}}}}\\
\Rightarrow{I_1} = 0.45 \times {10^4}{\rm{ A}}
\]
Let us write the expression for power loss during the step-down stage.
\[{P_1} = I_1^2R\]
Here \[{P_1}\] is the power loss during the step-down process.
On substituting \[0.45 \times {10^4}{\rm{ A}}\] for \[{I_1}\] and \[4.33{\rm{ }}\Omega \] for R in the above expression, we get:
\[
{P_1} = {\left( {0.45 \times {{10}^4}{\rm{ A}}} \right)^2}\left( {4.33{\rm{ }}\Omega } \right)\\
\Rightarrow{P_1} = 0.81 \times {10^6}{\rm{ W}}
\]
We cannot use this method for the transmission of power.
Let us write the expression for current when the voltage is stepped up using a step-up transformer.
\[{I_2} = \dfrac{P}{{{V_2}}}\]
On substituting \[{10^6}{\rm{ W}}\] for P and \[11000{\rm{ V}}\] for \[{V_2}\] in the above expression, we get:
\[
{I_1} = \dfrac{{{{10}^6}{\rm{ W}}}}{{11000{\rm{ V}}}}\\
\Rightarrow{I_1} = 90.909{\rm{ A}}
\]
We can write the expression for power loss during the step-up stage as below:
\[{P_2} = I_2^2R\]
Here \[{P_2}\] is the power loss during the step-up process.
On substituting \[90.909{\rm{ A}}\] for \[{I_2}\] and \[4.33{\rm{ }}\Omega \] for R in the above expression, we get:
\[
{P_2} = {\left( {90.909{\rm{ A}}} \right)^2}\left( {4.33{\rm{ }}\Omega } \right)\\
\Rightarrow{P_2} = 3.5 \times {10^4}{\rm{ W}}
\]
We know that the below relationship gives the fraction of power loss:
\[F = \dfrac{{{P_2}}}{P} \times 100\% \]
On substituting \[3.5 \times {10^4}{\rm{ W}}\] for \[{P_2}\] and \[{10^6}{\rm{ W}}\] for P in the above expression, we get:
\[
F = \dfrac{{3.5 \times {{10}^4}{\rm{ W}}}}{{{{10}^6}{\rm{ W}}}} \times 100{\rm{ }}\% \\
\therefore F = 3.578{\rm{ }}\%
\]
Therefore, the fraction of power loss during transmission is \[3.578{\rm{ }}\% \].

Note:The value of power loss obtained in the step-down stage is itself more than the power to be transmitted; therefore it is not feasible, and hence we did not consider it while reducing the fraction of power loss.