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
[tex]P_1 = \frac{V_1^2}{R}[/tex]
When the voltage is 1MV = 2 × 500kV
The power is
[tex]P_2 = \frac{V_2^2}{R}[/tex]
[tex]= \frac{2(V_1^2)}{R}[/tex]
[tex]= \frac{4V_1^2}{R}[/tex]
The ratio at which power is lost is
[tex]\frac{P_1}{P_2}[/tex][tex]= \frac{V^2_1}{R} * \frac{R}{4V_1^2}[/tex]
Here Resistance is constant
[tex]= \frac{1}{4}[/tex]
Hence the heat produced would be one fourth of the previous value
Answer:
[tex]\frac{P_1}{P_2} =\frac{1}{4}[/tex]
That means the heat loss wil be one fourth the previous value
Explanation:
Given that,
[tex]V_1 = 500kv\\\\V_2= 1MV\\\\V_2 = 2V_1[/tex]
The resistance R is the same in both cases
In the first case
The power loss is
[tex]P_1=\frac{V_1^2}{R}[/tex]
In second case
The power loss is
[tex]P_2=\frac{V_2^2}{R} \\\\P_2=\frac{(2V_1)^2}{R} \\\\P_2=\frac{4V_1}{R}[/tex]
Therefore, the ratio of the poer loss is
[tex]\frac{P_1}{P_2} =\frac{1}{4}[/tex]
That means the heat loss wil be one fourth the previous value
