Can you reduce amperage by using thicker wire with a constant voltage?
The amperage flowing through a wire is directly related to the load placed on the circuit, and has nothing to do with wire size, except that a larger wire will carry more amperage. Increasing wire size will not lower amperage but will allow the circuit to carry more amperage if the breaker is also increased in size.
No. Ohm's law tells us that V = IR. For a given load, R is constant, and thus the only way to reduce current is to increase voltage.
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Can you reduce amperage by using thicker wire or will this only avoid energy loss overheating and melting?
Normal household current is carried over a 10/2 wire. This is good to carry up to 20 amps. The more amps you wish to carry the larger the wire required. The smaller the number… the larger the wire. 8/2 is larger than 12/2. outside of the "standard room temperature" of the resistance of a given size of wire there is something called "heating losses" in any given size of wire ... simply put, the resistance of a given length of wire increases as it's temperature increases... drawing current thru a wire causes a certain degree (no pun intended) of heating, thus raising the resistance and lowering the voltage/current available to the load... normally (with the proper size of wire for the length of run) these effects (rt resistance/heating losses) are minimal. another name for the heating losses is "IRsquared loss" .
Answer . Voltage indicates the potential difference between two points. Ampere is the unit for current, indicating the magnitude of current. Voltage is due to which th…e current flows...and ampere is unit used for that current Voltage(V) and Ampere(A) can be related as follows R=V/I here I is the current(A) and v=voltage and R is termed as resistance Resistance is actually a opposing factor for the flow of current firstname.lastname@example.org
voltage is pressure amperage is number of elections per sec/
If you are referring to a simple circuit, you could add resistance throughout it. Increased resistance means decreased current flow yet the same voltage.
Yes. The current will vary depending on the varied resistive load placed on a fixed voltage. Lower the resistance, and current will increase.
Is there a chart available that will show the wire size amperage and voltage and distances that will be needed to safely wire a circuit?
You didn't specify the voltage that you are using but here is a simple easy to read chart that may meet your needs: http://www.fennelfamily.com/gti-vr6/electrical/cable-len…gth.html You can google "wire gauge distance" for more charts. Be careful and good luck!
It is mandatory to use a larger wire size to overcome voltage drop at the load.
As what i know, 30-50 Ampere per cathodes
As long as you don't exceed the current rating of the cable.
Amperage is the measure of electrical current, which is the measure of the electron flow through something (like a wire). The more electrons that flow through the wire, the h…igher the amperage. Current is understood as moving from higher voltage to lower voltage but since electrons are negatively charged, they actually flow in the opposite direction. Voltage is a measure of electrical potential between two items. The electrical potential can be looked at as the difference in the electrical charge between two items. The item with more negatively charged electrons has a lower voltage.
True or false if the voltage and resistance of a circuit stay constant the amperage can never change?
The equation for the three values in the question will give the definite answer. Amperage (I) is equal to the voltage (E) divided by the resistance (R). I= E / R So as you c…an see the answer is True. Example: 10 Volts and 50 Ohms in a circuit will have a current of .2 Amperes flowing through it. 10 / 50 = .2 You can also rearrange the equation to find the other two: E= R * E R= E / I
Transformers are used to increase the voltage and reduce the current in overhead power lines but wont increasing the voltage subsequently increase the current as resistance is a constant?
You're assuming that the line is dead shorted. In that case, assuming zero source impedance, current would increase as well. In reality, source impedance often limits the very… high voltage short circuit current to less than the lower voltage. Think of it this way: I have a 120 volt, 5000 watt totally resistive load (no motors). At 120 volts, I am pulling 41.67Amps. Say the power plant supplying this load is 100 miles away, and the overhead power line (regardless of voltage level) has .01 ohms/mile resistance (total of 1 ohm resistance). If the power company tries to deliver power at 120 volts, instead of the 5000 watts I want, I will get 5000 watts, but the power company will have to generate (5000 + 1 * 41.67^2) = 6736 watts*. If instead the power company steps the voltage up to say 13.8kV right outside my house (as close to the load as possible), total current at 13.8kV will be 362mA, so total power loss in transmission is .131watts (as opposed to 1736 at 120 volts). From the 120 volt perspective, my 5000 watt load "looks like" 2.88 ohms, since P = V*I = V^2/R. If we are looking at my house through a 13.8kV/120v transformer, the transformer has a turns ratio of 13800/120 = 115, thus increases voltage by 115 times, and decreases current by 115 times (from lowside to highside). Thus from the 13.8kV perspective, my 5000 watt load "looks like": P = V^2/R = (120 * turns_ratio)^2/R = (120*115)^2/R = 13800^2/R R = 38,088 ohms. The transformer changes voltage and current inversely to each other; this results in a change in apparent impedance relative to the highside and lowside of the transformer. *This is assuming the power company is delivering voltage at 120 volts through the line, and uses some sort of reactive power to compensate for the voltage drop through the line. This is often done by installing capacitor banks, or having generators closer to the load produce reactive power. The wasted transmission losses plus the cost of this extra equipment would result in higher power costs being passed on to customers.
Multiply (1,000) x (Voltage) x (Amperage) and you have KW. Multiply that by the number of hours you use it, and you have KWh.
Ampers are the ones that kill. But if there is too little voltage the electricity will not get through your body. On the other hand if theres 100000volts. and like 0.0000001 a…mps. You will also die because your heart cant take the shock.
Ohm's Law is Voltage = Resistance x Current So if you hold resistance constant, an increase in Voltage increases current by a proportional amount and vice versa if you decrea…se voltage.
To answer this question the resistance of the load is needed. I = E/R.
What is CLT wire and why can it with stand higher amperages and voltages than other wires types of the same size?
I researched CLT wire and found 2 possibilities. One is a heat trace cable and the other is a current limiting FLAT 2 conductor cable for 12 and 24 volt systems. I can only gu…ess since your question involves "higher amperages" than other conductors that you are referring to the heat trace cable. In either case, however, I believe the answer is the same. Different types of insulation can handle different amounts of current. The ampacity of a conductor is limited mostly by its insulation. This is to say that any BARE conductor in a certain application can carry more amps than it can with insulation. In most any case, the insulation melts and burns at temperatures below when the conductor itself melts. Glass or porcelain may be exceptions, as in fuses, but these insulators are not in contact with the conductors they protect. The voltage limit of a conductor is the maximum safe voltage an insulation can handle before the voltage breaches it, regardless of how many amps are flowing. The most common conductors in use today are rated for 600 volts. This rating means that applying more than 600 volts on the conductor could result in the voltage breaching (coming through) the insulation and allowing current to flow to any nearby conducting material, such as conduit or a panel enclosure. In other words it may be as if there is no insulation at all. This rating is mostly determined by insulation thickness, though material also plays a role.