0
\$\begingroup\$

In power distribution systems, a given amount of power can be transmitted with less voltage drop if a higher voltage is used. However,i don't understand the reason about this

In my thinking,if i use 110V(high voltage),the power can be transmitted with 10V (less voltage drop);However,if i use 11V(high voltage),the power can be transmitted with 10V (less voltage drop) too

It seems we can use less voltage drop to transmit the power if the "less" voltage is used too.Can anyone tell me about the mistake i do or explain the true meaning about "a given amount of power can be transmitted with less voltage drop if a higher voltage is used"

\$\endgroup\$
  • 2
    \$\begingroup\$ Lower current... \$\endgroup\$ – Brian Drummond Mar 26 at 12:39
1
\$\begingroup\$

Because the losses are due to the current squared times resistance.

So as the voltage increase, so the current reduces for the same resistance.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ OH,so the less voltage drop means less current because of V=IR with same resistance? \$\endgroup\$ – shineele Mar 26 at 12:39
  • 1
    \$\begingroup\$ Mike, I think you need to check / re-write your second paragraph. It's certainly not clear although, for some reason, the OP is happy! \$\endgroup\$ – Transistor Mar 26 at 13:43
0
\$\begingroup\$

Ignoring the difference between reactive power and resistive power for a minute...

Power P = V x A

So power in a load is the product of the voltage across it and the current flowing through it.

So if you want to dissipate 1 W in a resistor, you could deliver it in the following ways:

   Apply   1 V  to a      1 R  resistor, drawing  1.00 A
   Apply   5 V  to a     25 R  resistor, drawing  0.20 A
   Apply 100 V  to a  10000 R  resistor, drawing  0.01 A

The cable carrying the current to the load also has a resistance, let measure it in ohms/metre. The current through the cable will cause a voltage drop. That current, multiplied with the voltage drop it caused, will produce a power dissipation in the cable. And that power will heat it up.

The equation for the cable power is: Pc = I x I x Rc

So the less current through the cable, the less power dissipated in the cable.

In national power distribution, very high voltages are used to pass a now-smaller current through the cable and to a step-down transformer at the destination. The transformer, minus its inefficiency losses, will output a lower voltage but with higher current available.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

A simple example may help.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) Direct transmission at LV. (b) HV transmission.

  • In Figure 1a we drop 5 V on the feed and return wires giving a 10 V drop in voltage at the load. (These are all just rough numbers. The real voltage drop will be a little less because the current will be less than 100 A due to the added series resistance.) Power lost in the transmission is 2 × VI = 2 × 5 &times 100 = 1 kW. (The supply is 24 kW.)
  • In Figure 1b we drop only 0.5 V on the feed and return wires. We've stepped up the voltage by 10 so the line is running at 2400 V and the current falls to 10 A for the same power transmission. We now only drop 0.5 V per line, or 1 V total. When this drop is transformed back down through XFMR2 the voltage drop decreases to 0.1 V. Power lost in the transmission is 2 × VI = 2 × 0.5 &times 10 = 0.01 kW. (The supply is still 24 kW.)

In practice the user would typically trade some of the benefit by using lighter cable to save cost at the expense of poorer voltage regulation.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

Any time you transmit/transfer/send power (through wires in this case), not only is your transmitted power not all received/used at the end, but the voltage is reduced as well. So you have both a power loss and a voltage loss.
In order to transfer a given amount of power (let's say 100W), you can use any combination of voltage and current to get that much power, for example 10V at 10A, or 100V at 1A - the power is 100W in both cases. But the power delivered in the end is not 100W, and neither is the voltage the same at the end.

Here is a practical example: let's say you want to use the existing wiring in your house for the lights and use 10-volt bulbs instead of 100-volt bulbs, and let's say the wire is AWG 16 or around 1.3mm² and about 20 yards or 20 meters long from the power source to the bulb.
At 100V, the current will be 1A, the voltage drop on both wires will be around 0.53V (0.53% voltage drop), and the voltage across the bulb will be around 99.47V, which is unnoticeable to either you or your bulb (the bulb will be using almost 100W).

Reduce the voltage down to 10V, the current will have to be 10A for the same amount of power (100W), and now the voltage drop across the wires will be around 5.3V (53% or more than half of the voltage is dropped across the wires). The bulb will only have 4.7V across it and will use only 47W of the 100W sent from the power supply, which is definitely noticeable. The wires would get slightly warmer from the wasted heat and drop even slightly more voltage and power across them.

If you wanted to have the full 10V and 100W at the bulb, you would have to raise the power supply voltage to 15.3V (to make up for the voltage drop in the wires), but you would still be wasting the same amount of power, and you would have to supply 153W instead of 100W.
You could use much thicker wires, but they cost much more and are harder to work with. It would sometimes be hard to even fit them in the walls and other spaces.

P.S.: I have used 10V and 100V for the ease of calculation and understanding, and rounded some numbers, but it makes the point.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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