How does increasing the source voltage affect voltage drop and load voltage?

[torbor]

Does when why increase source voltage tu compensate losses why also increase voltage drop?. For example why increase current (reduce load resistance) and as side affect why increase voltage drop and then why increase source voltage tu counterbalance losses does also when why increase voltage for load why also increase voltage drop.

 

[Asymptotic]

Does when why increase source voltage tu compensate losses why also increase voltage drop?. For example why increase current (reduce load resistance) and as side affect why increase voltage drop and then why increase source voltage tu counterbalance losses does also when why increase voltage for load why also increase voltage drop

For linear ohmic materials, and ignoring temperature and other secondary effects, load resistance "is what it is".
Increasing current doesn't reduce load resistance.
Given a constant source voltage, however, current increases as resistance decreases and visa versa, because I=V/R.

What I believe you are asking is, "why does voltage drop increase (between point A and B in a circuit) when supply voltage increases?"
As supply voltage increases, what else increases? Current does.

As the amount of current flowing between points A and B increases, what happens to the voltage dropped between them? Because V=IR, and I has increased, so too will V, the voltage drop.

Sketching a simple multiple resistor series circuit, and fully working it out using Ohm's law will assist in understanding.

 

[torbor]

For linear ohmic materials, and ignoring temperature and other secondary effects, load resistance "is what it is".
Increasing current doesn't reduce load resistance.
Given a constant source voltage, however, current increases as resistance decreases and visa versa, because I=V/R.

What I believe you are asking is, "why does voltage drop increase (between point A and B in a circuit) when supply voltage increases?"
As supply voltage increases, what else increases? Current does.

As the amount of current flowing between points A and B increases, what happens to the voltage dropped between them? Because V=IR, and I has increased, so too will V, the voltage drop.

Sketching a simple multiple resistor series circuit, and fully working it out using Ohm's law will assist in understanding.

when load increase voltage drop also is increasing does when why increase source voltage to give load required voltage why also increase voltage drop

 

[cnh1995]

when load increase voltage drop also is increasing does when why increase source voltage to give load required voltage why also increase voltage drop

Does asymptotic's post not answer your question? Do you have something different in mind?

I see English is not your native language. Your exact question is not clear from your wording. You can use Google Translate. Type your question in your native national language (English font) and get it translated into English. Then copy-paste it here.

And, try to avoid using bold and bigger font as far as possible. It is considered as shouting.

 

[torbor]

Does asymptotic's post not answer your question? Do you have something different in mind?

I see English is not your native language. Your exact question is not clear from your wording. You can use Google Translate. Type your question in your native national language (English font) and get it translated into English. Then copy-paste it here.

And, try to avoid using bold and bigger font as far as possible. It is considered as shouting.

I really apologize for the bad english this is my question
When the current increases, the voltage drop also increases whether we increase the source voltage to give load required voltage Does this mean that we simultaneously increase the voltage drop

 

[cnh1995]

Does this mean that we simultaneously increase the voltage drop

Yes.

What I believe you are asking is, "why does voltage drop increase (between point A and B in a circuit) when supply voltage increases?"
As supply voltage increases, what else increases? Current does.

As the amount of current flowing between points A and B increases, what happens to the voltage dropped between them? Because V=IR, and I has increased, so too will V, the voltage drop.

 

[CWatters]

I think the situation you have is a non-ideal voltage source (with internal resistance Rs) connected to a load resistor (Rl)

The current drawn by the load causes a voltage drop across Rs and that reduces the voltage to the load.

If you increase the voltage of the source to compensate then the current flowing through Rs will also increase. This increases the voltage drop across the source resistance Rs. So the voltage of the source has to be increased slightly more than you might expect to produce the required output voltage.

Does this mean that we simultaneously increase the voltage drop

Yes it all happens simultaneously.

 

[jim hardy]

Does when why increase source voltage tu compensate losses why also increase voltage drop?.

It seems to me torbor is asking the 'why' behind Kirchoff's Voltage law.

The answer comes from the reasoning we learned in grade schoool geometry class - "two things that are equal to the same thing are equal to each other."

Both Vs_ource and Vl_oad are equal to Vobserved , so they must be equal to each other.

@torbor
Here's how to work Kirchoff's Voltage Law

Imagine yourself very small and walking inside the circuit around the path i drew in red.

Write down each voltage you encounter along the way, with its sign which is the first one you meet. There's no voltage drop along wires so just keep walking until you encounter the next component with voltage.
Walking the path starting at lower left corner, up, over and down to lower right corner, I get
-Vs +Vl
When you get back to where you started write down "= 0" . Then look at what you wrote.
-Vs + Vl = 0
That is Kirchoff's Voltage Equation for the circuit above.
You will find it is always true. That's why it's a law.

Re-arranging it gives Vl = Vs .

That i think is the answer to your question.

You MUST teach yourself this discipline of walking the circuit and writing Kirchoff's Voltage Law every time..
It will become intuitive, as it has for everyone who's tried to answer your question before me.
When it does "click" with you, you will soon be unable to remember when it didn't seem as natural as tying your shoe.

Just form the habit. We all had to .

old jim

 

[jim hardy]

Hmmm i should have drawn my red path down through the load box

if one wishes he could take it through Vobserved and write that into the Kirchoff equation
which would be a good exercise for torbor, write simultaneous equations to prove both Vs and Vl are equal to Vobserved.

That's how you gain confidence in yourself.

I still remember that day in 2nd grade when i first tied a shoelace by myself. I did it about four times just to relish the success.

 

[adaliadella]

Does when why increase source voltage tu compensate losses why also increase voltage drop?. For example why increase current (reduce load resistance) and as side affect why increase voltage drop and then why increase source voltage tu counterbalance losses does also when why increase voltage for load why also increase voltage drop.

1. Increasing the number or size of conductors.
2. Reducing the load current on the circuit.
3. Decreasing conductor length, and.
4. Decreasing conductor temperature.

 

[Tom.G]

also when why increase voltage for load why also increase voltage drop

When the voltage at a fixed load increases the current increases. (Currrent = Voltage / Resistance) If you do not change the wire size, the increased current flowing thru the wires from the supply causes a bigger voltage drop. (Volts = Current x Resistance)

 

[Babadag]

In my opinion, whatever you try to say is how by raising the supply voltage at far end of a line you’ll get elevated voltage at the load terminals since so doing we would increase the voltage drop across the supply line.
At first, the load current is not always direct propotional with the voltage-if the load is an induction motor-for instance-the current could even decrease.
Second, usually the line impedance is less than 10% of the load impedance and the increasing rate of supply voltage will be divided according to impedance: more for the load and less for the line.
For instance, let's take a 2 kW heater [pure resistance-approximate] 230 V single phase 100 m supply cable of 2*2.5 mm^2 copper conductor. The cable resistance-for this cross section area the reactance is negligible- is 100/58/2.5 at 20oC and for 75 oC a factor of (234.5+75)/(234.5+20)=1.216 it has to be employed. Total resistance 2*100/58/2.5*1.216=1.68 ohm [forward and backward].
The rated current is 2000/230=8.7 A
The load resistance is 230^2/2000=26.45 ohm
The current now will be 230/(1.68+26.45)=8.18 A
The cable voltage drop will be 1.67*8.18=13.66 V
Then the voltage at load terminal will be 230-13.66=216.34 V
The heater will deliver only 8.18*216.34=1.7697 kW
Now we shall increase the supply voltage by 5% [230*1.05=241.5 V]
The new current will be 241.5/(26.45+1.67)=8.59 A
The new voltage drop on cable will be 1.67*8.59=14.35 V
However, the voltage at load terminal will be 241.5-14.35=227.2 V
The heater will deliver 8.58*227.2=1.949 kW