0
The power dissipated by a 10 ohm resistor with 800v across it is 64 kw.
Ohm's law: current is voltage divided by resistance
Power law: power is voltage times current, so power is voltage squared divided by resistance
Don't even think about trying this. 64 kw is a lot of power. The resistor will probably explode, or catch fire. At best, the 80 amps required will trip your circuit breaker, if you are lucky.
What else can I help you with?
What process will double the power dissipated by a resistor?
Increase the voltage across the resistor by 41.4% .
What is the power dissipated by a 1.2 Kilohm resistor if the voltage drop across the resistor is W volts?
The power dissipated by a resistor can be calculated using the formula ( P = \frac{V^2}{R} ), where ( P ) is the power, ( V ) is the voltage drop across the resistor, and ( R ) is the resistance. For a 1.2 kilohm resistor (or 1200 ohms), the power dissipated would be ( P = \frac{W^2}{1200} ) watts. Thus, the power dissipated depends on the square of the voltage drop across the resistor divided by 1200.
If the voltage dropped across a resistor increases by a factor of 10 the power dissipated by the resistor is?
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.
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
A 100 ohm resistor is placed across a 100v power source what is the power?
Power dissipated in a resistance = E2/R = (100)2/100 = 100 watts.
What process will double the power dissipated by a resistor?
Increase the voltage across the resistor by 41.4% .
What is the power dissipated by a 1.2 Kilohm resistor if the voltage drop across the resistor is W volts?
The power dissipated by a resistor can be calculated using the formula ( P = \frac{V^2}{R} ), where ( P ) is the power, ( V ) is the voltage drop across the resistor, and ( R ) is the resistance. For a 1.2 kilohm resistor (or 1200 ohms), the power dissipated would be ( P = \frac{W^2}{1200} ) watts. Thus, the power dissipated depends on the square of the voltage drop across the resistor divided by 1200.
How much power will be dissipated by a 9 ohm resistor if 9 volts are applied across it?
P = (E2)/R = 81/9 = 9 watts
If the voltage dropped across a resistor increases by a factor of 10 the power dissipated by the resistor is?
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.
How Determine the power dissipated by 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).
What is the power dissipated by a resistor with a current of 2 amps and a resistance of 1000 ohms?
I = 2A R = 1000Ω Power Dissipated P = I2R = (2A)2(1000Ω) = 4000W Voltage across resistor V = IR = (2A)(1000Ω) = 2000V
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
A 100 ohm resistor is placed across a 100v power source what is the power?
Power dissipated in a resistance = E2/R = (100)2/100 = 100 watts.
What is the voltage drop running through resistor 1 resistor 1 equals 3 ohms?
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.
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.
If two resistors of equal value are connected in parallel with each other. The power dissipated by the two resistors will be equal to the sum of the power dissipated by each resistor true or false?
True. When two resistors of equal value are connected in parallel, the total power dissipated by the circuit is indeed the sum of the power dissipated by each resistor. Since they have the same resistance and are subjected to the same voltage, each resistor will dissipate the same amount of power, and their combined power will equal twice that of one resistor.
Voltage drop of 15 Volts across 2.7 killo ohm resistor?
You just stated that the voltage across the resistor is 15 volts, so that's your answer ! If the resistor is connected to a 15-V battery or to the output of a 15-V power supply, then a meter across the resistor is also across the power supply, and reads 15 volts. The current through the resistor is (V/R) = (15/2700) = 5.56 mA. The power dissipated by the resistor (and delivered by the battery) is (V2/R) = (225/2700) = 0.083 watt.
