Question

If the direct transmission method with a cable of resistance $0.4\,\Omega {\text{k}}{{\text{m}}^{ - 1}}$ is used, the power dissipation (in %) during transmission is
A. 20
B. 30
C. 40
D. 50

Answer 1

Hint
The question is based on the relation between the power, resistance, current and voltage. We can calculate the resistance of the cable then we can calculate the value of power input using the voltage and current value and output power again using the current and the resistance value. The power dissipated will be the ratio of input by output power.

Complete step by step answer
A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers' usage.
Therefore, the conductivity of a cable is, $0.4\Omega {\text{k}}{{\text{m}}^{ - 1}}$and distance between the power plants is 20 km. Therefore, the net, that is, the total resistance of the cable used for the data transmission is given as,
$\Rightarrow R = 0.4{\Omega \mathord{\left/
 {\vphantom {\Omega {km}}} \right.} {km}} \times 20\,km$
Hence on calculating the resistance we get,
$\Rightarrow R = 8\Omega $
The amount of the electric power is given to be 600 kW and the voltage supplied is of amount 4000 V for the cable used for the data transmission. Now, we will compute the amount of the current flow through this cable.
From the formula for power we can calculate the current using the value of the voltage,
$\Rightarrow P = VI$
Substituting the values we get,
$\Rightarrow 600 \times {10^3} = 4000 \times I$
So to get the value of current,
$\Rightarrow I = \dfrac{{600 \times {{10}^3}}}{{4000}}$
On calculating this we get,
$\Rightarrow I = 150\,A$
Therefore, the amount of the current flow is 150 A.
We have just obtained the amount of the current flow through the cable used for the data transmission and its resistance. Now, we can compute the amount of power transmitted through this cable by the formula,
$\Rightarrow P = {I^2}R$
Substituting the values,
$\Rightarrow P = {150^2} \times 8$
On calculating we get the power as,
$\Rightarrow P = 180\,kW$
Therefore, the amount of power is 180 kW
The loss per cent is calculated as follows,
$\Rightarrow {\text{loss}} = \dfrac{{180}}{{600}} \times 100\% $
So on calculating we get the loss as,
$\Rightarrow {\text{loss}} = 30\% $
Therefore, the power dissipation (in %) during transmission is 30.
Hence, option (B) is correct.

Note
The other method, a direct method, also called the direct transmission method can also be applied to solve this problem as follows,
$\Rightarrow {\text{Power}}\,{\text{loss}} = \dfrac{{{\text{loss}}}}{{{\text{input}}}} \times 100\% $
Now we can substitute the corresponding equations in the above expression and we get,
$\Rightarrow {\text{Power}}\,{\text{loss}} = \dfrac{{{\raise0.7ex\hbox{${{V^2}}$} \!\mathord{\left/
 {\vphantom {{{V^2}} R}}\right.}
\!\lower0.7ex\hbox{$R$}}}}{{600\,k}} \times 100$
Substituting the values we get
$\Rightarrow {\text{Power}}\,{\text{loss}} = \dfrac{{{{4000}^2}}}{{8 \times 600 \times {{10}^3}}} \times 100\% $
On calculating this value is,
$\Rightarrow {\text{Power}}\,{\text{loss}} = 30\% $