1 coulomb is the electric charge carried in 1 second by a current of 1 amp. 1 joule is the work done in one second at a power of 1 watt. So they are different things, with different dimensions, you can't equate them. To find the power caused by a flow of 1 coulomb you have to know the resistance of the circuit, the power in watts is then I2R.
The volt is a derived unit of electrical potential. It is equal to joules per coulomb, or kilogram meter squared per ampere second cubed.
An ampere is a unit of charge flow rate, while a watt is a unit of energy flow rate. The two units are not directly convertible. More specifically, one ampere is one coulomb per second, while one watt is one joule per second. If you knew how many volts were involved, you could compare, because volts is joules per coulomb.
To calculate the energy stored in a battery with volts and coulombs, you can use the formula: Energy (Joules) = Voltage (Volts) x Charge (Coulombs). Multiply the voltage by the charge of the battery to get the energy capacity in Joules.
148 calories is 619.6 joules. Since one calorie = 4.18 joules then you multiply that number by 148 and you get 618.64 joules (619 rounded)
The answer is 0.001 kilojoules in a joule 1J = 0.001kJ
1 volt = 1 joule per coulomb 3 joules x 1 coulomb = 3 volts
it would be 10 joules because all you do is divide 10 joules by 1 coulomb of charge and you get 10 joules or (V) volts
In a 120-volt circuit, each coulomb of charge flowing receives 120 joules of energy. This is calculated using the formula: energy = voltage x charge. So, 120V x 1C = 120 joules.
119 joules per coulombCharges don't get joules as they flow through a circuit. They lose them.Every coulomb of charge that flows through a circuit ... from one terminal of a119-volt power supply, around the circuit, and back to the other terminal ...loses 119 joules during the trip.
24 volts (joules per coulomb), alternating current
No way of telling. to get amps you have to have a current flow, which you get when you connect a consumer to an outlet. Then the consumer will pull amps according to its wattage rating (Watts / Volts = amps) - assuming it's all hooked to a fuse with enough rating.
A joule/coulomb is represented by the volt. Example: a 9v battery provides 9 joules of energy to every coulomb of charge that passes through it.
A Coulomb is the unit of charge. It is a fundamental unit, representing the number of elementary charges (typically, electrons) available to do work. Its numerical value is about 6.241510x1018 elementary charges Important combined units based on the coulomb are the ampere, which is coulombs per second, the volt, which is joules per coulomb, and the volt-ampere, which is joules per second, or watts.
12 watts or 12 joules of energy.In one second, 1 coulomb is 1 amp, so the power is 1 amp x 12 volts = 12 watts, and in that one second, that is 12 Joules of energy.
A single lightning strike typically releases about 1 billion joules of energy.
Voltage is electromotive force, in joules per coulomb. Power is energy transfer rate in joules per second, also known as watts.Not asked, but answered for completeness sake, and also to show the relationship between voltage and power, current is charge transfer rate in coulombs per second. So, if you multiply voltage (joules per coulomb) by amperes (joules per second) you get watts (joules per second).
That depends on the voltage. In general, a coulomb of charge will either gain or lose (depending on the direction) one joule of energy for every volt of potential difference. For example, if the battery has 12 V, a coulomb of charge will gain or lose 12 joules of energy when going from one terminal to the other.