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".
The power dissipated in a resistor is I2R = (0.5)2 (5) = 0.25 x 5 = 1.25 watts
5 volts applied to 50 ohms will induce a current of 0.1 amperes. (I = E/R) 5 volts applied to a current of 0.1 amperes will dissipate 0.5 watts. (P = I*E)
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 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 !
A current limiter. Commonly used in well designed power supplies to make the power supply "Fold over" if output current exceeds a safe value. Commonly use a low value resistor in series with the output, and a circuit that senses the voltage across it. Although the above answer describes a circuit to limit current, the device that limits the amount of current flowing through it is a resistor.
66W 230/800= 0,2875. 230= 66w
Electrical power is measured in watts. The formulas for power dissipated by a resistor are: * P = V × I * P = V² / R * P = I² × R P is power, V is voltage, I is current, R is resistance.
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.
.205 watts or 205 mw
Power dissipated = I2R 0.022 x 1000 = 0.4 watts
I = E / RIf the voltage across the resistor is 90 volts, and the resistance of the resistoris 9 ohms, then the current through the resistor is90/9 = 10 Amperes.Don't try this at home!The power dissipated by the resistor is E2/R = (90)2/9 = 900 watts. That's comparable to the power (heat) dissipated by a small toaster. A common composition resistor will get hot and possibly explode if it's asked to dissipate that kind of power.
A typical resistor will burn out when it dissipates power in excess of double its power dissipation rating for an extended period of time. The power dissipated by a resistor is equal to I2R or E2/R, where E = the voltage across the resistor I = the current through the resistor R = the resistance of the resistor
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).
I = 2A R = 1000Ω Power Dissipated P = I2R = (2A)2(1000Ω) = 4000W Voltage across resistor V = IR = (2A)(1000Ω) = 2000V
Due to energy usage and/or the reduction in conductance (increase in resistance) in a given load or resistor, some electrical energy is lost through that component. As such, a proportional drop in current and voltage occurs.
P = I^2 x R] P = 0.2^2 x 100 P = 4 W
Depends on the current. Put a resistor in-line with the current, then measure the voltage across the resistor. V=RI. So, divide the measured voltage by resistor value. Be careful with the size of the resistor, as Power dissipated in a resistor is R*I^2 or V^2/2. So, a 1-Amp current into a 1 Ohm resistor will result in a 1Watt power dissipated in the resistor. If it's too small, it'll burn. Also, notice that if you do that, you haven't measured the current in the original circuit. You've measured the current when an extra resistor is installed in the original circuit, and that's different.
Power = I2 R = (0.02)2 x (1,000) = 0.4 watt
Increase the voltage across the resistor by 41.4% .