66W
230/800= 0,2875. 230= 66w
The question has just stated clearly that the applied voltage is 12 volts DC.Provided that the power supply is capable of maintaining its output voltage while supplying some current ... i.e. that the effective internal resistance of the power supply is small ... and that the 2.7 ohm resistor is the only external element connected to the power supply's output, the voltage across the resistor is exactly 12 volts DC.The current through the resistor ... supplied by the 12 volt DC supply ... is 12/2.7 = 4.44 Amperes (rounded).The power dissipated by the resistor ... supplied by the DC supply ... is 122 / 2.7 = 53.23 watts !
If by power supply you mean a voltage source, it really won't matter that the resistor is removed. The voltage source will provide infinite current, instantly charging the capacitor so that the capacitor's voltage is equal to the source.Alternative AnswerIf you are referring to an a.c. circuit, then a load current will continue to flow with its value being determined by the capacitive reactance of the circuit, and the resulting phase angle will lead the supply voltage be very close to 90 degrees.
The power in a resistor (in watts) is simply the product of the current (in amperes) times the voltage (in volts).The power in a resistor (in watts) is simply the product of the current (in amperes) times the voltage (in volts).The power in a resistor (in watts) is simply the product of the current (in amperes) times the voltage (in volts).The power in a resistor (in watts) is simply the product of the current (in amperes) times the voltage (in volts).
The current I = 0.18257 amperes. Scroll down to related links and look at "Electrical voltage V, amperage I, resistivity R, impedance Z, wattage P".
Current = (voltage) / (resistance) = 110/20 = 5.5The current will be 5.5 Amperes RMS, alternating at 50 Hz.Note:If you try this at home, you must use a gigantic "power" resistor.The resistor dissipates E2/R = I2R = 605 watts RMS !
You just stated that the voltage across the resistor is 15 volts, so that's your answer ! If the resistor is connected to a 15-V battery or to the output of a 15-V power supply, then a meter across the resistor is also across the power supply, and reads 15 volts. The current through the resistor is (V/R) = (15/2700) = 5.56 mA. The power dissipated by the resistor (and delivered by the battery) is (V2/R) = (225/2700) = 0.083 watt.
7.0w/4 = 1.75w
Increase the voltage across the resistor by 41.4% .
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.
The question has just stated clearly that the applied voltage is 12 volts DC.Provided that the power supply is capable of maintaining its output voltage while supplying some current ... i.e. that the effective internal resistance of the power supply is small ... and that the 2.7 ohm resistor is the only external element connected to the power supply's output, the voltage across the resistor is exactly 12 volts DC.The current through the resistor ... supplied by the 12 volt DC supply ... is 12/2.7 = 4.44 Amperes (rounded).The power dissipated by the resistor ... supplied by the DC supply ... is 122 / 2.7 = 53.23 watts !
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).
A 120V power supply connected to a 30 Ohm resistor will produce 120/30 or 4 amps of current.
.205 watts or 205 mw
In a DC circuit, the power dissipated by a resistance is (voltage across it)2 divided by 'R'.P = E2/R = (14.1)2 / 142 = 198.81/142 = 1.4 watts(rounded)
I = 2A R = 1000Ω Power Dissipated P = I2R = (2A)2(1000Ω) = 4000W Voltage across resistor V = IR = (2A)(1000Ω) = 2000V
Power dissipated = I2R 0.022 x 1000 = 0.4 watts