0
$\begingroup$

Is the Ohm's law violated here? And also why is electricity generated at 11,000 V in India?

$\endgroup$
  • $\begingroup$ This is a repost of another question by the same user. $\endgroup$ – DanielSank Nov 20 '17 at 6:52
  • 1
    $\begingroup$ @DanielSank You got it the wrong way around - this here is the older question, the other is the repost. $\endgroup$ – ACuriousMind Nov 20 '17 at 12:54
2
$\begingroup$

Is the Ohm's law violated here?

No, the Ohm's law is not violated since Ohm's law relates the voltage across ('between the ends') of a conductor to the current through the conductor.

But the voltage on a transmission line (the voltage referenced to ground) is not the voltage across the transmission line (the voltage difference between the ends of the transmission line).

To apply Ohm's law correctly requires understanding which voltage and current are related by it.

As given in another answer, for a given power, we require that the product of the transmission line voltage and current is constant. So, holding the power constant, an increase in voltage results in a decrease in current.

Since the ohmic losses in a transmission line are proportional the square of the current, reducing the current by 50% cuts the power loss by 75%.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The current required to carry a given power decrease when you increase the voltage because the power is the product of the current with the voltage (and power factor).

| cite | improve this answer | |
$\endgroup$
-3
$\begingroup$

as per the formul;I=P/V......SO if we keep the value of 'P' CONSTANT AND VARRY THE VALUE OF 'V'then we get the value of i will less.on the other hand ohms law applicable only in a particular conductor's two end......and should have fix impedance.In transmission line we are using transformer and it doesn't have fixed impedance becoz we always doing stepup and step down to the power.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.