What is the formula for calculating power (watts) - Weebly

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What is the formula for calculating power (watts)

Watts (W) - volts (V) - amps (A) - ohms (次) calculator. Calculates dc power / voltage / current / resistance. Enter 2 values to get the other values and press the Calculate button: Ohm's law calculator ? Ohms calculations The resistance R in ohms (次) is equal to the voltage V in volts (V) divided by the current I in amps (A): The resistance R in ohms (次) is equal to the squared voltage V in volts (V) divided by

the power P in watts (W): The resistance R in ohms (次) is equal to the power P in watts (W) divided by the squared current I in amps (A): Amps calculations The current I in amps (A) is equal to the voltage V in volts (V) divided by the resistance R in ohms (次): The current I in amps (A) is equal to the power P in watts (W) divided by the voltage V in volts (V): The current I in amps (A) is equal to the square

root of the power P in watts (W) divided by the resistance R in ohms (次): Volts calculations The voltage V in volts (V) is equal to the current I in amps (A) times the resistance R in ohms (次): The voltage V in volts (V) is equal to the power P in watts (W) divided by the current I in amps (A): The voltage V in volts (V) is equal to the square root of the power P in watts (W) times the resistance R in ohms (次):

Watts calculation The power P in watts (W) is equal to the voltage V in volts (V) times the current I in amps (A): The power P in watts (W) is equal to the squared voltage V in volts (V) divided by the resistance R in ohms (次): The power P in watts (W) is equal to the squared current I in amps (A) times the resistance R in ohms (次): Ohm's law calculator ? See also Power is the rate of using or supplying

energy: Power is measured in watts (W)Energy is measured in joules (J)Time is measured in seconds (s) Electronics is mostly concerned with small quantities of power, so the power is often measured in milliwatts (mW), 1mW = 0.001W. For example an LED uses about 40mW and a bleeper uses about 100mW, even a lamp such as a torch bulb only uses about 1W. The typical power used in mains

electrical circuits is much larger, so this power may be measured in kilowatts (kW), 1kW = 1000W. For example a typical mains lamp uses 60W and a kettle uses about 3kW. Power = Current ℅ Voltage There are three ways of writing an equation for power, current and voltage: where: P = power in watts (W)V = voltage in volts (V)I = current in amps (A) or: P = power in milliwatts (mW)V = voltage in volts

(V)I = current in milliamps (mA) You can use the PIV triangle to help you remember these three equations. Use it in the same way as the Ohm's Law triangle: To calculate power, P: put your finger over P, this leaves I V, so the equation is P = I ℅ V To calculate current, I: put your finger over I, this leaves P over V, so the equation is I = P/V To calculate voltage, V: put your finger over V, this leaves P over I,

so the equation is V = P/I The amp is quite large for electronics so we often measure current in milliamps (mA) and power in milliwatts (mW). 1mA = 0.001A and 1mW = 0.001W. Using Ohm's Law V = I ℅ R we can convert P = I ℅ V to: where: P = power in watts (W) I = current in amps (A) R = resistance in ohms () V = voltage in volts (V) You can use triangles to help with these equations too: Normally

electric power is useful, making a lamp light or a motor turn for example. However, electrical energy is converted to heat whenever a current flows through a resistance and this can be a problem if it makes a device or wire overheat. In electronics the effect is usually negligible, but if the resistance is low (a wire or low value resistor for example) the current can be sufficiently large to cause a problem. You

can see from the equation P = I? ℅ R that for a given resistance the power depends on the current squared, so doubling the current will give 4 times the power. Resistors are rated by the maximum power they can have developed in them without damage, but power ratings are rarely quoted in parts lists because the standard ratings of 0.25W or 0.5W are suitable for most circuits. Further information is

available on the resistors page. Wires and cables are rated by the maximum current they can pass without overheating. They have a very low resistance so the maximum current is relatively large. For further information about current rating please see the cables page. The amount of energy used (or supplied) depends on the power and the time for which it is used:A low power device operating for a long

time can use more energy than a high power device operating for a short time. For example: A 60W lamp switched on for 8 hours uses 60W ℅ 8 ℅ 3600s = 1728kJ. A 3kW kettle switched on for 5 minutes uses 3000W ℅ 5 ℅ 60s = 900kJ. The standard unit for energy is the joule (J), but 1J is a very small amount of energy for mains electricity so kilojoule (kJ) or megajoule (MJ) are sometimes used in scientific

work. In the home we measure electrical energy in kilowatt-hours (kWh), often just called a 'unit' of electricity when the context is clear. 1kWh is the energy used by a 1kW power appliance when it is switched on for 1 hour: For example: A 60W lamp switched on for 8 hours uses 0.06kW ℅ 8 = 0.48kWh. A 3kW kettle switched on for 5 minutes uses 3kW ℅ 5/60 = 0.25kWh. You may need to convert the kWh

domestic unit to the scientific energy unit, the joule (J): 1kWh = 1kW ℅ 1 hour = 1000W ℅ 3600s = 3.6MJ This website does not collect personal information. If you send an email your email address and any personal information will be used only to respond to your message, it will not be given to anyone else. This website displays advertisements, if you click on these the advertiser may know that you came

from this site and I may be rewarded. No personal information is passed to advertisers. This website uses some cookies classed as 'strictly necessary', they are essential for operation of the website and cannot be refused but they do not contain any personal information. This website uses the Google AdSense service which uses cookies to serve advertisements based on your use of websites (including

this one) as explained by Google. To learn how to delete and control cookies from your browser please visit . ? John Hewes 2021 Learn the Power Formula We*ve seen the formula for determining the power in an electric circuit: by multiplying the voltage in ※volts§ by the current in ※amps§ we arrive at an answer in ※watts.§ Let*s apply this to a circuit example: How to

Use Ohm*s Law to Determine Current In the above circuit, we know we have a battery voltage of 18 volts and a lamp resistance of 3 次. Using Ohm*s Law to determine current, we get: Now that we know the current, we can take that value and multiply it by the voltage to determine power: This tells us that the lamp is dissipating (releasing) 108 watts of power, most likely in the form of both light and heat.

Increasing the Battery*s Voltage Let*s try taking that same circuit and increasing the battery*s voltage to see what happens. Intuition should tell us that the circuit current will increase as the voltage increases and the lamp resistance stays the same. Likewise, the power will increase as well: Now, the battery*s voltage is 36 volts instead of 18 volts. The lamp is still providing 3 次 of electrical resistance to the

flow of current. The current is now: This stands to reason: if I = E/R, and we double E while R stays the same, the current should double. Indeed, it has: we now have 12 amps of current instead of 6. Now, what about power? What does Increasing a Battery*s Voltage do to Power? Notice that the power has increased just as we might have suspected, but it increased quite a bit more than the current. Why

is this? Because power is a function of voltage multiplied by current, and both voltage and current doubled from their previous values, the power will increase by a factor of 2 x 2, or 4. You can check this by dividing 432 watts by 108 watts and seeing that the ratio between them is indeed 4. Using algebra again to manipulate the formula, we can take our original power formula and modify it for applications

where we don*t know both voltage and current: If we only know voltage (E) and resistance (R): If we only know current (I) and resistance (R): Joule*s Law Vs. Ohm*s Law A historical note: it was James Prescott Joule, not Georg Simon Ohm, who first discovered the mathematical relationship between power dissipation and current through a resistance. This discovery, published in 1841, followed the form of

the last equation (P = I2R), and is properly known as Joule*s Law. However, these power equations are so commonly associated with the Ohm*s Law equations relating voltage, current, and resistance (E=IR ; I=E/R ; and R=E/I) that they are frequently credited to Ohm. REVIEW: Power measured in watts, symbolized by the letter ※W§. Joule*s Law: P = I2R ; P = IE ; P = E2/R RELATED WORKSHEETS: Try

out our Ohm*s Law Calculator in our Tools section.

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