Question

When electrical power is transmitted over large distances, it is economical to use a high voltage and low
current to minimize the I2 R heating loss in the transmission lines. Transformers are useful for
transmitting power over long distances because when the generator voltage is stepped up, the
current in the transmission lines is reduced, thereby reducing I2 R losses. A generator produces 10 A
(rms) of current at 400 V. the voltage is stepped up to 4500 V by an ideal transformer and
transmitted a long distance through a power line of total resistance 30  .
With the help of the comprehension given above, choose the most appropriate alternative to each of
the following questions.
1. The percentage of power lost in the transmission line when the voltage is stepped up is :
(A) 10% (B) 20%
(C) 0.60% (D) 2%
2. The percentage of power lost in the transmission line if the voltage were not stepped up is :
(A) 100% (B) 90%
(C) 75% (D) 50%
3. If the transmission line is cooled so that the resistance is reduced to 5.0  and it carries a
current of 0.89 A, the power lost in the line is :
(A) 5 W (B) 4 W
(C) 8 W (D) 24 W

Answer 1

Voltage and current of primary and secondary circuit of transformer:

$$\begin{gathered} I_1=10\mathrm{A} \\ {V}_1=400\mathrm{V} \\ {V}_2=4500\mathrm{V} \\ {I}_{\mathit{2}}\mathit{=}\frac{{\mathit{V}}_{\mathit{1}}{\mathit{I}}_{\mathit{1}}}{{\mathit{V}}_{\mathit{2}}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\frac{\mathit{400}\mathit{\times }\mathit{10}}{\mathit{4500}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\frac{\mathit{8}}{\mathit{9}}\mathrm{A} \end{gathered}$$

(1)

Percentage of power lost when voltage is stepped up:

$$\begin{gathered} \%ofP=\frac{{I_2}^2{R}_2}{V_1{I}_1}\times 100 \\ =\frac{{\left({\displaystyle \frac{8}{9}}\right)}^2\left(30\right)}{\left(400\right)\left(10\right)}\times 100 \\ =0.6\% \end{gathered}$$

(2)

Percentage of power lost when voltage were not stepped up:

$$\begin{gathered} \%ofP=\frac{{I_1}^2{R}_1}{V_1{I}_1}\times 100 \\ =\frac{{\left({\displaystyle 10}\right)}^2\left(30\right)}{\left(400\right)\left(10\right)}\times 100 \\ =75\% \end{gathered}$$

(3)

Power lost:

$$\begin{gathered} △P=I^{2}R \\ ={\left(0.89\right)}^2\left(5\right) \\ =4\mathrm{W} \end{gathered}$$