Voltage times current. You obtain current from the division of voltage and resistance, so:
I[A] = U[V] / R[ohm]
and
P[W] = U[V] * I[A]
it follows, that
P[W] = U[V] * (U[V] / R[ohm]) = U[V] ^ 2 * R[ohm]
So, voltage squared divided by resistance will give you the power that will be dissipated in a resistor. Whether the resistor will take that abuse is up to its power dissipation rating, however.
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,
so say y = 2^x dy/dx = ln2.2^x (. = multiplication symbol, ^ = to-the-power-of symbol) The general formula is (where 'a' is a constant, x is what you are differentiating with respect to and y is f(x)) y = a^x then dy/dx = lna.a^x Go ask a math teacher or look up exponential function differentiation on the internet for why.
FV of growing annuity = P * ((1+r)^n - (1+g)^n) / (r-g) P=initial payment r=discount rate or interest rate g=growth rate n=number of periods ^=raised to the power of NB: This formula breaks when r=g due to division by 0. When r=g, use P * n * (1+r)^(n-1)
integral of e to the power -x is -e to the power -x
x times x to the first power is x to the second power
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.
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 formula P = I^2R relates power (P), current (I), and resistance (R), indicating the power dissipated in a resistor. On the other hand, the formula P = V^2/R relates power (P), voltage (V), and resistance (R), representing the power dissipated across a resistor. The former formula deals with power in terms of current, while the latter formula expresses power in terms of voltage.
.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
P = I^2 x R] P = 0.2^2 x 100 P = 4 W
Power dissipated = I2R 0.022 x 1000 = 0.4 watts
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 rating of a resistor determines how much power it can dissipate without being damaged. For example, a 1/4W resistor is designed to handle up to 1/4W continuously without being destroyed. When selecting a resistor to use in a circuit, use Ohm's law to calculate the power it will dissipate. For example, placing a 1kΩ resistor across a 12VDC signal will allow 12/1000 = 0.012A to flow thru the resistor. 0.012*12 = 0.144W will be dissipated. Thus, a 1/8W (0.125W) resistor would not be sufficient, and a 1/4W (0.25W) must be used.
You may find it helpful to use Ohm's law and the definition of electrical power.