Here are a few easy and related equations.
V = Voltage, I = Current (amps)
R = Resistance, (Ohms)
V = I X R
R= V / I
I = V / R
Power = Watts = W, is expressed as V X I. Since V =I X R, W then also = R X I X I
Which is I **2 X R
Same as V**2 / R
So, if your I = 3 and your R = 3, and because I X R = V,
3 X 3 =9 Volts
(Note: capacitance and inductance can also influence the amount of current flow in a system. This resistance is referred to as reactance. Both reactances are a function of the frequency of the Voltage applied.)
Have a wonderful day. Bob
Increase the voltage across the resistor by 41.4% .
Facts missing. Need two of three values. The resistors value. The voltage across it. The current flowing through it. P=I^2*R. P=E^2/R
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
There is insufficient information in the question to answer it. You need to provide either the voltage across the resistor, or the power dissipated by the resistor. please restate the question.
real power (as opposed to imaginary power, which is not dissipated)
The formula for calculating the power dissipated in a resistor, known as the i2r power, is P I2 R, where P is the power in watts, I is the current in amperes, and R is the resistance in ohms.
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
A series circuit has 100mA flowing through a 1.5kohm load. The power dissipated by the load is equivalent to 15 Watt. This is based on the formula, power is equals to square current times load.
Increase the voltage across the resistor by 41.4% .
Facts missing. Need two of three values. The resistors value. The voltage across it. The current flowing through it. P=I^2*R. P=E^2/R
The power dissipated in a voltage divider circuit is given by the formula P = V^2/R, where V is the voltage across the resistor and R is the resistance of the resistor. If the resistance in the voltage divider circuit is increased, the power dissipated in the circuit will decrease. This is because as resistance increases, the current flowing through the circuit decreases, leading to less power being dissipated as heat in the resistors.
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
There is insufficient information in the question to answer it. You need to provide either the voltage across the resistor, or the power dissipated by the resistor. please restate the question.
real power (as opposed to imaginary power, which is not dissipated)
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.
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).