v(81v - 100)
The power dissipated by a resistor is given by the formula ( P = \frac{V^2}{R} ), where ( V ) is the voltage across the resistor and ( R ) is its resistance. If the voltage increases by a factor of 10, the new power can be expressed as ( P' = \frac{(10V)^2}{R} = \frac{100V^2}{R} = 100P ). Therefore, the power dissipated by the resistor increases by a factor of 100.
only the iMac and the mac Pro
why not ? a 1000kva means you will get the power out as 1000kv per amp in essence it can be a 100v with 10 amp out. means the same thing to me. It's power factor depends upon the load. Thus we can not give absolute value of KW rating of a Xmer or an alternator in the manufacturing plant.
100v2 - 220v + 121 = (10v - 11)2
Power dissipated in a resistance = E2/R = (100)2/100 = 100 watts.
100 V2 - 25 W2 = (10V + 5W) (10V - 5W)Another way to show it:100 V2 - 25 W2 = 25 (4 V2 - W2) = 25 (2V - W) (2V + W)
To find the current required by a 75W bulb at 100V, you can use the formula ( P = V \times I ), where ( P ) is power, ( V ) is voltage, and ( I ) is current. Rearranging the formula gives ( I = \frac{P}{V} ). Plugging in the values, ( I = \frac{75W}{100V} = 0.75A ). Therefore, the bulb requires 0.75 amperes of current.
LM317 can be used for regulated dc power supply upto 37V. For more voltage range, a zener can be connected at the reference pin.
0.1Uf 100v
Power = Voltage * CurrentIsolating current, we getCurrent = Power/Voltage, I = 800W/100V = 8 amps
To convert 110V to 100V, you can use a step-down transformer. A step-down transformer reduces the voltage while maintaining the same power output. Simply plug the device requiring 100V into the step-down transformer, and it will adjust the voltage accordingly.
100v divided by 1.41