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The peak voltage is 1.414 times the rms.

V=I*R Where: V is voltage I is the current in ampers R is resistance in ohms. So, if the current is 15 A and the resistance is 5 ohms, then the voltage must be 15 A *5 ohms = 75 V.

By Ohm's Law, current is voltage divided by resistance, so a voltage of 60 volts across a resistance of 15 ohms would yield a current of 4 amperes.

current = voltage/ resistance = 1.5/15 =0.1 A

Voltage dividers can provide anything between VCC (the most positive voltage in the circuit) and VSS (the most negative voltage in the circuit). For example, if VCC =0 and VSS = -15, then the output voltage has to be negative. ========================

15 by 67 that will give you the actual amount

Ohm's Law states that V=I*R. In this case, I=15 Amps and R=8 Ohms. The voltage therefore will be 120 volts.

at the 15 side of the coil. 15 is ignition on voltage

You can use a 8x AGP card but the motherboard will throttle it down to 4x, so you lose any performance gain you would get with a AGP 8x motherboard. 8x cards are cheap, and 4x cards are hard to find, so it's not going to hurt anything to buy something more powerful than what your motherboard will run it at. Pricewise, an 8x card goes from $15 to $50 for low-end graphics.

Change is from 15 to 15 = 25 - 15 = 10 units. 10 units as a percentage of 15 units = 10/15*100 = 66.67 %

This is a voltage drop question. A voltage must be stated to answer this question.

That depends on the voltage you use. If there are 120 volts, the power equals voltage times amperage. 120 volts times 15 amps = 1800 watts. Cheers ebs

A #14 copper wire with an insulation factor of 90 degrees C is rated at 15 amps. To answer your question for voltage drop at 200 feet a voltage needs to be stated. Assuming the voltage of 120 is used to maintain 15 amps at the distance of 200 feet a #6 copper conductor will limit the voltage drop to less that 3 percent. Assuming the voltage of 240 is used to maintain 15 amps at the distance of 200 feet a #10 copper conductor will limit the voltage drop to less that 3 percent.

This is a voltage drop question. To answer this question a voltage has to be stated. The higher the voltage to the circuit becomes the smaller the wire size needed. After a certain voltage point the wire size will remain constant and the voltage drop at the load will become smaller.

% of change = [(21-15)/15] x 100%= 40% (increase)

This is a voltage drop question. A voltage of the load has to be stated. State the voltage under the discussion tab and the question will be answered.

It depends on the voltage; which depends on which country you live in.

At what voltage? P equals I times E, Where p=power in watts, I=current in amps and E= voltage in volts.

It depends on the voltage. I = P / E where I = amps, P = watts, E = voltage. Examples: 12 volts (such as in a car): I = 15/12 I = 1.25 amps 120 volts (like a small lamp in a US home): I = 15/120 I = 0.125 amps

Is the voltages divided by the resistance 60/15=4 amps

The voltage should be in the range of 13.2 to 14 or so volts. It should not go over 15 volts.

The terminal strip's rating is 15 amps at 600 volts. It does not matter what the voltage is up to 600 volts, the maximum amperage allowed on the strip is 15 amps. It could be 15 amps at 12 volts or 15 amps at 600 volts or any voltage in between.

A voltage regulator converts a/c volatge from the stator to dc voltage and regulates it down to under 15 volts. When they go bad they can over charge and take out a battery as well as not charge at all.

what is the chemical change

I work for the company.Their are may factors.Time in the company,Your title(if you work on High Voltage lines and are your title)Currently they start @ 15 and change an hour.Thats with you having no title and in training.The max pay for a high voltage troubleshooter is upper 30s or low 40s an hr.

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