Discuss Voltage drop. in the UK Electrical Forum area at ElectriciansForums.net
say you start with a 240V supplty and a load of 100 ohms. I = V/R = 240/100 = 2.4AThanks guys.
So is my understanding right.
Voltage drop along a conductor (VXI=P) so if said voltage was reduced due to voltage drop. The new voltage is proportional to current so as the voltage has been reduced so will the current causing the overall power to also be reduced?
Revisit my post #5 above. Sometimes it decreases, sometimes stays nearly constant, sometimes increases, depending on the nature of the load.so will the current causing the overall power to also be reduced?
But for the kettle example for the resistive loaf the current has gone up so the power is kept as a constant so as voltage drops current increases due to loss in voltage.Revisit my post #5 above. Sometimes it decreases, sometimes stays nearly constant, sometimes increases, depending on the nature of the load.
No, that is the wrong way round. For the kettle example which is a basic resistor then the total loss will always be proportionally split between it and the cable, but as the supply voltage increases both increase.So the voltage drop ultimately leads to Power losses in the Kettle and more current to be drawn in the conductor.
So a 2200w kettle will have have more power at a higher voltage? As per example they are both 22 ohms resistance so identical. So in a practical example if we buy a kettle which states that as it’s load and we use it on a 230v supply it will not be as powerful as it would using a 250v supply?No, that is the wrong way round. For the kettle example which is a basic resistor then the total loss will always be proportionally split between it and the cable, but as the supply voltage increases both increase.
For example, if your cable is 1 ohm and kettle is 22 ohms then total is 23 ohms and at 230V you have 230V / (1R + 22R) = 10A.
Cable loss = I^2.R = 10 x 10 x 1 = 100W
Kettle power = I^2.R = 2200W
Ratio = 2200 / 100 = 22 (as for original resistance)
Repeat for 250V and I = 250/23 = 10.67A
Cable loss = 10.67 x 10.67 x 1 = 113.8W
Kettle power = 10.67 x 10.67 x 22 = 2505W
Ratio = 2505 / 113.8 = 22 (to rounding)
Again it is obvious it will always be that as (I^2.Rcable) / (I^2.Rload) = Rcable/Rload
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For a restive load it is straightforward to compute the other case, when you have a fixed supply and add cable resistance because you have:
I = V / (Rcable + Rload)
For other loads as Lucien Nunes has pointed out there is no fixed 'Rload' value but it depends on the supply voltage.
In that case the old method of solving this would be to plot the load-line (i.e. current versus voltage) of the load, and the the slope of the supply (voltage against current), and where they intersect is your operating point (i.e. combination of current and voltage that simultaneously satisfies both the characteristic of the supply impedance and that of the attached device).
Exactly. For any simple restive load like a heater the power is proportional to the voltage squared:So a 2200w kettle will have have more power at a higher voltage? As per example they are both 22 ohms resistance so identical. So in a practical example if we buy a kettle which states that as it’s load and we use it on a 230v supply it will not be as powerful as it would using a 250v supply?
Thanks for your help!Exactly. For any simple restive load like a heater the power is proportional to the voltage squared:
I = V / R
P = V x I
P = V x (V / R) = V^2/R
The current is always the same no matter where you measure it. It is determined by the overall resistance of the circuit from end to end. For an individual device in the circuit the voltage drop depends on the resistance of that single device (Ohms Law : V = IR) and the sum of all the individual voltage drops in the circuit is equal to the total voltage accross the circuit from end to end. That's Kirschoff's Voltage Law.Can someone clear up a few points for me please. With voltage drop when the voltage is decreased by a % will the current increase to make up for the loss in voltage to give us the same output?
And the more current we draw the greater the volt drop?
Thank you
Going off on a tangent, this is why the DNO typically likes to keep the supply voltage as high as they can. A small change in voltage can cause a disproportionate change in I^2 R losses in their network - which is power they have to pay for but can't sell.Also important aspect is for a fixed cable resistance (i.e. typical case from your design) the power loss goes up with the square of the load current which can become expensive over a long period.
But for a inductive reactive load such as a motor this is false as the voltage decrease the current increases?Exactly. For any simple restive load like a heater the power is proportional to the voltage squared:
I = V / R
P = V x I
P = V x (V / R) = V^2/R
If you are dealing with a fixed shaft power output then yes, the V x I should be more or less constant. But depending on what a motor is driving and the specific characteristics of the motor the current may well go up with voltage.But for a inductive reactive load such as a motor this is false as the voltage decrease the current increases?
But for a inductive reactive load such as a motor this is false as the voltage decrease the current increases?
Thank you for your explanation, when would the motor current increase when the voltage falls? Is there a practical example?Motor current can increase as the voltage falls, but that is not because it is inductive.
A pure inductance is linear and passes a current proportional to voltage. I=V/xC = V/2.f.L The nearest approximation to a pure inductance found in normal electrical work would the be the primary of an unloaded transformer. Within reason, the magnetising current is proportional to the voltage, but as it lags the voltage by 90° it absorbs no power. In any real transformer there are losses even when unloaded so there is also a power component of the current which, although it increases with increasing voltage, is non-linear as it depends on the magnetic characteristics of the iron core which are also non-linear.
Motors much more complicated; different kinds of motors with different kinds of loads behave in very different ways. To expand on PC1966's comment about motors in which the speed is determined mainly by the frequency (synchronous, induction) and those where the torque is nearly constant, consider a hoist, where the torque depends on the weight being lifted, not the speed it is moving. The power component of the current is approximately proportional to torque, but inversely proportional to flux density (from the general motor characteristics.) Since flux density increases with voltage (although not linearly) the effect of voltage drop in the cable would be to increase the power component of current but decrease the wattless magnetising component. The power factor would improve and there would be some arbitrary variation in current magnitude.
As he points out, a motor of non-constant speed, especially when coupled to a load with a steep torque/speed curve, is likely to show decreased power and current due to voltage drop. Therefore it is impossible to make the generalisation: 'With increasing voltage, motor current varies as...." For a more thorough explanation we would have to delve into some AC circuit maths and get into motor characteristic equations etc.
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