You may find it helpful to use Ohm's law and the definition of electrical power.
Increase the voltage across the resistor by 41.4% .
Power dissipated in a resistance = E2/R = (100)2/100 = 100 watts.
real power (as opposed to imaginary power, which is not dissipated)
No. This would mean more power is used than is provided - an impossibility. It is possible that you may not be accounting for a power source in your circuit, or have the actual source modelled incorrecty.
Power dissipated = I2R 0.022 x 1000 = 0.4 watts
No, because the power dissipated in a resistor is proportional to the square of the current through the resistor but only directly proportional to the resistance of the resistor (I^2 * R) and the current through the lower value resistor will be higher than the current through the higher value resistor, the lower value resistor will usually dissipate more power.
Increase the voltage across the resistor by 41.4% .
Power dissipated in a resistance = E2/R = (100)2/100 = 100 watts.
1amp
real power (as opposed to imaginary power, which is not dissipated)
Power dissipated by the resistor = I^2 * R or V^2 / R, where R = its resistance value, I = the current in the resistor, and V = the voltage drop across the two terminals of the resistor. You need to measure or find the information of either I (using an ammeter) or V (a voltmeter).
The effective resistance of those three resistors in parallel is 20 ohms. And it makes no difference what the power source is, or whether they're even connected to a power source at all. As soon as those three resistors are in parallel, their effective resistance is 20 ohms immediately, even if they're still in the drawer.
6
You need a conductor, power source and a resistor. You need a conductor, power source and a resistor. You need a conductor, power source and a resistor.
No. This would mean more power is used than is provided - an impossibility. It is possible that you may not be accounting for a power source in your circuit, or have the actual source modelled incorrecty.
.205 watts or 205 mw
I = 2A R = 1000Ω Power Dissipated P = I2R = (2A)2(1000Ω) = 4000W Voltage across resistor V = IR = (2A)(1000Ω) = 2000V