Physics 2 student here. I’ve just learned that:
P=I*V
Using ohms law, V=I*R you can get the equations:
P=I^(2)*R and P=V^(2)/R
In these equation, power is proportional to I^2 and V^(2) respectively. This seems to show that power is squarely proportional to both voltage difference and current. Yet it’s stated that high voltage results in less power loss than high current. How can this be true when power scales by the same amount with a change in either.
The context of this is power lines and why they use high voltage.
Please ELI5. I’ve heard inadequate explanations in my textbooks, I can’t wrap my head around it.
In: Physics
I^2*R “power” is typically shown as heat loss in a conductor. A big part of why circuit breakers in your home are sized (15A, 20A) is due to the heat generated based on Current and Resistance in your home wiring.
So! If you can knock down the current, and ramp up the voltage (think power grid), you can transmit power more efficiently.. with less heat loss.
Perhaps that is what the book is trying to say.
Because the power lost in the wires as heat is I^2 *R, so higher current means hotter wires. Resistance increases with temperature as well which makes it worse.
One way to think of it, V = IR gives you the voltage drop across a resistor R from current I. Smaller I means a smaller voltage drop across that resistor, which would lower V^2 /R, the power transmitted through the resistor. Remember in these equations, V is not the voltage the resistor in the circuit is at, but rather the voltage change across it.
For an overall circuit, P = IV where I is the current and V is the source voltage – ground. But that’s a different power from what is just being consumed by the resistance in the wires.
Power loss is the resistance in the wires multiplied by the current.
Double the voltage and you halve the current, and therefore the power loss is half.
In particular, you have to consider that while current I is a fixed amount, ***voltage V is relative.*** The P = V^(2)/R formula is based on the voltage drop **across the device.**
In other words, let’s throw out the low-voltage/high-voltage thing entirely. Simple circuit: V = 10 V source. Two 5 Ohm resistors in series. Total resistance is 10 Ohms, so current is 10 V / 10 Ohms = 1 A. Hopefully you agree with this.
The power loss through the first resistor is P = I^(2)R = (1 A)^(2)(5 Ohms) = 5 W. Same for the second. This is a total power of 10 W, which again agrees with what we expect (P = VI = 10 V x 1 A = 10 W).
If we use the voltage form of P = V^(2)/R, we **do not** calculate the first resistor’s power as (10 V)^(2)/(5 Ohms) = 100 / 25 W = 25 W. We have to use the **voltage drop across the resistor** which is only 5 V to get (5 V)^(2)/(5 Ohms) = 25 / 5 W = 5 W.
Going back to the low-voltage/high-voltage case: my bet is that you are incorrectly thinking that the V used in the V^(2)/R calculation is the input voltage. As we just saw **it is not.** It is the **voltage drop.** So the answer is that when you go from a lower voltage to a higher voltage with the same power, current necessarily drops (P = VI => V increase, P constant -> I decrease); which means that the voltage drop V = IR decreases, and the voltage on the other side of the line increases.
Voltage drop = voltage this side – voltage that side. Increasing voltage that side means that the voltage drop ***decreases***. So yes, V^(2) increases from the fact that the voltages are being increased, but it also **decreases** because of the reduced voltage drop across a resistive load from a lower current and the net effect is that losses are reduced.
Side note: this is a reason to work primarily in terms of current when analyzing these kinds of situations. Voltage analysis gets really tricky quite quickly, because you have to think in terms of the voltage drops.
Power loss on a transmission line is given by I^2 *R.
So minimizing current (I) results in minimizing power losses.
The reason for this is that wires get hot when you run current through them. The more current you run through them, the hotter they get. That heat represents energy that has been lost from the electrical system. (resistance also tends to increase with heat, making high currents a double whammy).
I think the misunderstanding lies is that with P = V^2/R the voltage here is the difference in voltage at then ends the wire, not comared to neutral. So while you double the voltage, the difference at the outer ends of the wire is not doubled.
The power loss along a conductor is due to resistance. This manifests as a voltage drop along the conductor. So which equation to use depends on what you’re able to measure or control.
V = IR can tell us how much voltage is lost along the conducting path. This turns into a power loss because P=I*R, so P=I^2 *R.
P=V^2 /R gets us the same answer if we’ve measured the voltage drop and the path resistance.
Where this gets interesting for your question is to explore power lost from a fixed amount of delivered power (e.g. 10000W) and fixed resistance (e.g. 0.1ohm) for various combinations of current and voltage.
0.1ohms with 100V and 100A at the load means 100V+(100A^2 *0.1ohm)=1100V at the input.
Bump that up to 1000V and 10A at the load, 1000+(10^2 *0.1)=1010V at the input.
That “V” in V^2/R is the voltage difference between the ends of a wire. It is not the rated or nominal voltage.
Say you have a 9 ohm load supplied by a 1 ohm wire. Nominally you supply 10V to the circuit. What happens is that 1V drop is seen in the wire and 9V over the load. Effectively the load outputs 9^2/9 = 9 W and 1^2/1 = 1W is lost as heat in the wire.
Say you want to deliver the same power at higher efficiency. The load is increased to 899 ohms and the wire remains at 1 ohm. Instead you use 90V supply. Now the voltage drop is 89.9V on the load and only 0.1V on the wire. The power at the load is 8.99W and the loss in the wire is 0.01W. So the efficiency now is increased delivering the same power to the load.
Two things that I haven’t seen anyone mention yet is that 1) we use I²R for power loss because current is what causes things to heat up, not voltage; and 2) across a piece of wire that has virtually 0Ω, you’re not going to find any voltage drops to be able to calculate V²/R, but you will find current with which to calculate I²R.
For 1), think about it. Say you have a series capacitor on a simple DC circuit. If you let it charge over time, eventually the capacitor will act like an open circuit (this is its steady-state response), since it basically is just two metal plates with a non-conducting medium in the center. Once it acts like an open circuit, current will no longer be flowing through the circuit, but the voltage across the capacitor will be the same as the power source. You’ll have, I dunno, 12V across it, but no charges will be moving anywhere. Is it consuming power if there is no movement of charges? Will anything heat up if there is no current of electricity releasing energy? No, it won’t be consuming power and thus won’t be heating up, even though its voltage is 12V. Once you turn off the power supply and the capacitor starts discharging, and current starts flowing through the circuit again, THEN there will be power consumption again. This is also why having a AAA battery in your hand doesn’t cause it to heat up; it’s not doing any work, which means it’s not using any energy, and thus there is no energy dissipation, no power is being used.
For 2), you know that power is measured in Watts, which is the same as Joules/second. When charges are moving, they are using energy to move. The unit of energy is Joules. As charges move, they interact with their medium and transfer some of their energy into it, which is then released as heat. If too many charges are flowing through, then the heat released increases. Current is just the movement of electric charges, so a higher current means that you have a higher number of electric charges in motion, and the more electric charges you have in motion, the more energy they transfer to the wire, which gets released as heat. Resistance is the capacity of something to resist the movement of charges across it. Generally, the higher the resistance of a component, the more it resists the movement of charges across it. Because charges need a larger amount of energy to be able to “punch through” the resistive component, you won’t see it consume power/heat up until there is enough difference in charges across the component (or voltage), in which case the current will “punch through” and it will light the component on fire instantly, as too much energy was used at once. Therefore, a good measure of power is then I²R.
Actually when you step up voltage you reduce resistance by square of that factor so less resistance means obviously less power loss.
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