Question

If a power utility were able to replace an existing 500 kV transmission line with one operating at 1 MV, it would change the amount of heat produced in the transmission line to four times the previous value. two times the previous value. one-fourth of the previous value. none of these. The amount of heat produced would remain the same. one-half of the previous value.

Answer 1

Complete Question

If a power utility were able to replace an existing 500 kV transmission line with one operating at 1 MV, it would change the amount of heat produced in the transmission line to

a four times the previous value.

b two times the previous value.

c one-fourth of the previous value.

d The amount of heat produced would remain the same.

e one-half of the previous value.

f none of these.

Answer:

Option C is the correct answer

Explanation:

When the voltage is 500kV

    The power is

              $P_1 = \frac{V_1^2}{R}$

When  the voltage is 1MV = 2 × 500kV

    The power is

            $P_2 = \frac{V_2^2}{R}$

                 $= \frac{2(V_1^2)}{R}$

                 $= \frac{4V_1^2}{R}$

The ratio at which power is lost is

                   $\frac{P_1}{P_2}$$= \frac{V^2_1}{R} * \frac{R}{4V_1^2}$

Here Resistance is constant

                     $= \frac{1}{4}$

Hence the heat produced would be one fourth of the previous value

                     

       

Answer 2

Answer:

$\frac{P_1}{P_2} =\frac{1}{4}$

That means the heat loss wil be one fourth the previous value

Explanation:

Given that,

$V_1 = 500kv\\\\V_2= 1MV\\\\V_2 = 2V_1$

The resistance R is the same in both cases

In the first case

The power loss is

$P_1=\frac{V_1^2}{R}$

In second case

The power loss is

$P_2=\frac{V_2^2}{R} \\\\P_2=\frac{(2V_1)^2}{R} \\\\P_2=\frac{4V_1}{R}$

Therefore, the ratio of the poer loss is

$\frac{P_1}{P_2} =\frac{1}{4}$

That means the heat loss wil be one fourth the previous value