R stands for the resistance.
R = r*L/A where: R = resistance r = material resistivity L = length A = area Since you are doubling the diameter, you are increasing the area by a factor of 4, so Y will have 1/4 the resistance of X.
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
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.
To calculate the resistor needed to pull 1500 mA (1.5 A), you need to know the voltage across the resistor. Using Ohm's Law (V = I × R), you can rearrange it to find R (Resistance = Voltage / Current). For example, if the voltage is 5V, the resistor needed would be R = 5V / 1.5A = 3.33 ohms. However, ensure the resistor can handle the power dissipation, calculated as P = I² × R, which in this case would be approximately 7.5 watts.
The resistance of a resistor will be numerically equal to its resistivity when the resistor's dimensions (length and cross-sectional area) are such that the resistance ( R ) is calculated using the formula ( R = \frac{\rho L}{A} ) and the units align accordingly. Specifically, this occurs when the length ( L ) of the resistor is equal to its cross-sectional area ( A ). In such a case, ( R ) would equal ( \rho ) if they are expressed in compatible units.
resistor-thermistor
p=I*I*R ,P=V*V/R;where I is the current passing through the resistor, and V is the voltage across resistor, and R is the Resistance of the resistor,
the R stands for Resistor. It is a resistor plug.
what does resistor mean in technolagy
it's generally the resistance. its generally the energy dissipiating component called 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).
R = r*L/A where: R = resistance r = material resistivity L = length A = area Since you are doubling the diameter, you are increasing the area by a factor of 4, so Y will have 1/4 the resistance of X.
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
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.
To calculate the resistor needed to pull 1500 mA (1.5 A), you need to know the voltage across the resistor. Using Ohm's Law (V = I × R), you can rearrange it to find R (Resistance = Voltage / Current). For example, if the voltage is 5V, the resistor needed would be R = 5V / 1.5A = 3.33 ohms. However, ensure the resistor can handle the power dissipation, calculated as P = I² × R, which in this case would be approximately 7.5 watts.
Resistor Relay
Resistor