0
What is the power dissipated by an 8.2k ohm resistor if a current of 0.005 amps is passing through the resistor?
Wiki User
∙ 14y ago
Take your pick:
P = V x I (Power = Voltage x Current)
or:
P = V2 / R (Power = Voltage2 / Resistance)
or:
P = I2 *R (Power = Current2 x Resistance)
(the last two equations come from combining the ohms law equation R=V/I with the power equation P=VxI)
In the question above you have resistance and current therefore:
P = I2 *R = 0.0052 x 8.2k = 0.0052 x 8200 = 0.205W = 205mW
Wiki User
∙ 14y ago
Add your answer:
Earn +20 pts
2 resistors unequal value are in parallel. Would the power dissipated by the resistor with larger ohmic value be greater than the power dissipated by the resistor of lesser value?
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.
What is the functions of resistor?
It slows down the speed of the electrons , which are all passing through the resistor
What are the effects of power on a resistive device?
Current passing through a resistor, et al, causes heat. The heat causes the resistance of said resistor to decrease, which causes current to increase, and the cycle just keeps going until the circuit burns out.
The no-load output voltage of a DC power supply is measured at 15V When a 600 ohms load is connected to the output the output drops to 13.7V Calculate the internal resistance of the power supply?
This question can be answered using voltage dividers. Assume the power supply consists of a voltage source and a resistor. With no load, all of the voltage source's voltage is dissipated by the internal resistor of 15V. When there is a load, there are two resistors in series. To calculate the internal resistance:1. I=V/R. You know the 600ohm resistor dissipated 13.7V. So that would mean a current of 13.7/600=22.8mA2. If the 600ohm resistor dropped 13.7, kirchoff's voltage law would tell us the internal resistor dropped 15-13.7=1.3V.3. R=V/I, Use the current to calculate the internal resistance. 1.3/22.8mA = 56.9ohmsCommentFurther to the above answer, a voltage-source's voltage is not 'dissipated by the internal resistance when on no load'. On no load, there is no current passing through the internal resistance, so no 'voltage dissipation' can takes plac -i.e. the non-load voltage is 15 V.
Is the heat loss and current of a resistor affected by being in a parallel circuit or can you just calculate it the same as in series?
The heat generated by any particular resistor depends (at least electrically) solely on the power it dissipates. Power dissipation in a resistor is equal to current squared times resistance, and the current through the resistor is equal to the voltage across it divided by the resistance. If we take a 10 ohm resistor ('your resistor') and put it in a series circuit such that there is 10 volts across your resistor, the current through it will be 1 ampere (10/10=1). the power dissipated will be 10 watts (1^2 * 10=10). If we put your resistor in a parallel circuit that also puts 10 volts across it, then the current and power will be the same. Your resistor does not know or care where the voltage came from. From this point of view, once you get down to the voltage across the resistor, it does not matter what type of circuit it is in. On the other hand, for any given power supply voltage, then the type of circuit and the value of external components certainly does affect the terminal voltage and thus the current through as well as the power dissipated by the resistor. In a parallel circuit, the voltage across your resistor remains basically the same no matter what resistance you put in parallel with it (unless you overload the power supply or the power supply has high internal resistance). In this case, the voltage across the resistor is the same as the power supply, current is I=E/R, R being that resistor only, and power is P=I^2 * R. In a series circuit the current through the resistors is I=E/R, R being the total resistance (including the other resistor(s)). The power dissipation in your resistor will then be P=I^2 * R, I being the series current we just calculated, and R being your resistor only. Since the other resistors affect the current, and since the current is the same no matter where you measure in a series circuit, then the voltage across your resistor and thus the power dissipation will be affected. The voltage across your resistor will be E=I*R, I being the series current we just calculated, and R being your resistor only. So, while the calculation for power dissipated in a particular resistor does not change relative to what type of circuit it is in, the calculation to arrive at the voltage across the resistor and/or the current through it (which you will then need to calculate power) does. Keep in mind there are other mechanical parameters that influence the actual case temperature of the resistor. Physical size of the case, composition, and airflow velocity, if any, will alter the case-to-ambient thermal conductivity. Ambient temperature will also be a factor in the final temperature.
2 resistors unequal value are in parallel. Would the power dissipated by the resistor with larger ohmic value be greater than the power dissipated by the resistor of lesser value?
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.
What power is dissipated by an 8.2 k resistor if a current of 0.005 amps pass's through the resistor?
.205 watts or 205 mw
What is a ballast resistor?
A ballast resistor is an electrical resistor whose resistance varies with the current passing through it, thus maintaining a constant current.
Find the power dissipated by an 8.2k resistor if a current of 0.005 amps is passing through the resistor?
0.205 W Power is equal to voltage times current (P=V*I) in Watts, and from Ohm's Law, the voltage across a capacitor is equal to the current passing through it times the resistance (V=I*R). The power is then P=I2R. For a current of 5mA and a resistance of 8.2kOhms, (5*10-3A)2*(8.2*103Ohms) = 0.205 W.
What are the colors of a resistor and why are they there?
There are many colours for resistor rings but the reason that we have them is to help the resistor reduce the electrical current that is passing through.
Is an resistor an conductor or a insulator?
A resistor is a conductor that dissipates some of the electrical energy fromthe current flowing through it. The energy dissipated by the resistor is(current through it)2 x (resistance)
How do you find power generated in a resistor?
The power generated in a resistor is converted into heat. and that can be power which is converted into heat is the product of the voltage across the resistor and, current passing through the resistor. or the product of square of the current and the resistance offered by the resistor.
What is a functoin of a resistor?
esistors restrict the flow of electric current, for example a resistor is placed in series with a light-emitting diode (LED) to limit the current passing through the LED.
What is the current running through a resistor of 9 ohms with volts of 90?
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.
At what level will typical resistors burn out?
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
Why you cannot measure the resistance of a resistor when there is current passing through it?
you can if you know the current measure the voltage across the resistor E=I*R there is actually no way to measure the resistance without passing current thru it P=I*E E=I*R any 2 will give you the other 2
How are power dissipated in a load or in a resistor?
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.
