Question about P=V*I, P=R*I^2 and V=R*I (electricity)

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In summary, the conversation discusses the relationship between voltage, current, and power in high voltage power lines. It is explained that in order to reduce power loss, the voltage is increased and the current is reduced. However, there is confusion about the different values of power and how it relates to Ohm's law. The experts clarify that there is a difference between the power delivered to the load and the power lost in transit, and that the goal is to minimize the I^2*R loss on the transmission line. A diagram is provided to further illustrate the concept. Overall, the conversation provides a clear understanding of the relationship between voltage, current, and power in high voltage power lines.
  • #1
zooflute
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This is my first post, and hopefully it's in the right section (my apologies if not!)...I have posted this on Answers, but there weren't any helpful responses there, so I though this might be a better crowd to ask.

Ok, we were taught in our science (electricity) class that in order to reduce power loss (P) on high voltage power lines, they increase the voltage (V) and reduce the current (I), which makes P go down by the square of I. So with this formula:

1) If V goes up by 10, then I goes down by 10, since P=VI
So if I goes down by 10, then P goes down by 100, since P=R*I^2

But this doesn't make sense when you put it into Ohm's law:

2) If V goes up 10 and R is constant, then I goes up 10. Using V=I*R
So if I goes up by 10 then P goes up by 100, since P=R*I^2

Also, in 1), why are the two "P"s different? (in P=V*R, P should be constant, but in the P=R*I^2 part, P goes up 100) Are they different types of P?

Any insight on this we be really appreciated!
 
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  • #2
First, you should distinguish between the power being delivered to the load
and power being 'wasted' in the power distribution line.

The power delivered to the load, call it Pd, is V*I at the receiving end.
The power lost in transit, your P, is the I^2*R loss on the line.

The idea is to reduce the I^2*R loss on the transmission line for a given constant
power Pd being delivered to the load. The I^2*R loss on the line is accompanied
by an end-to-end voltage loss on the line: DV = I*R. Note that DV is not V.

A careful analysis will reveal that it is a tad more complicated, because the voltage
delivered at the receiving end is actually V - DV, so the power delivered to the load
will really be (V - DV)*I. We want DV to be small compared to V.
 
  • #3
It's nice to see a good question and a good answer come from two first time posters.
 
  • #4
Good explanation from gneill
This diagram may help to see what's happening.
The figures are just chosen for illustration.
Cable has 10 ohm resistance, supply has 1000W power.
It could be supplied at, for example, 1000V x 1A (top) or 500V x 2A (bottom)
powerloss.png
 
  • #5
Wow thank you both so much! Your answers put together helped me understand perfectly.
Good stuff ^^
 

FAQ: Question about P=V*I, P=R*I^2 and V=R*I (electricity)

Question 1:

What is the relationship between power (P), voltage (V), and current (I) in electricity?

The relationship between power, voltage, and current is described by the equation P=V*I, where P is power in watts, V is voltage in volts, and I is current in amps. This equation is known as Ohm's law and shows that power is directly proportional to both voltage and current.

Question 2:

What is the difference between P=R*I^2 and V=R*I in terms of electric circuits?

P=R*I^2 and V=R*I are two different forms of Ohm's law. P=R*I^2 is known as the power law and is used to calculate the power dissipated in a circuit when the resistance (R) and current (I) are known. V=R*I is known as the voltage law and is used to calculate the voltage drop across a resistor (V) when the resistance (R) and current (I) are known.

Question 3:

How do P=R*I^2 and V=R*I relate to each other?

P=R*I^2 and V=R*I are both derived from Ohm's law and are used to describe different aspects of an electric circuit. P=R*I^2 is used to calculate power dissipated, while V=R*I is used to calculate voltage drop. However, both equations are related as they both involve the same variables (R and I) and are used to describe the flow of energy in a circuit.

Question 4:

Can P=R*I^2 and V=R*I be used interchangeably in calculations?

No, P=R*I^2 and V=R*I cannot be used interchangeably in calculations. These equations are used to describe different aspects of an electric circuit and should be used accordingly. For example, P=R*I^2 would be used to calculate power dissipated in a circuit, while V=R*I would be used to calculate the voltage drop across a resistor.

Question 5:

What are the units for power (P), voltage (V), and current (I) in P=R*I^2 and V=R*I?

The units for power, voltage, and current in P=R*I^2 and V=R*I are as follows: P (power) is measured in watts (W), V (voltage) is measured in volts (V), and I (current) is measured in amps (A). It is important to use the correct units when performing calculations to ensure accurate results.

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