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Increase the voltage across the resistor by 41.4% .
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.
The power dissipated by a 10 ohm resistor with 800v across it is 64 kw.Ohm's law: current is voltage divided by resistancePower law: power is voltage times current, so power is voltage squared divided by resistanceDon'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.
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.
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
Increase the voltage across the resistor by 41.4% .
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.
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).
I = 2A R = 1000Ω Power Dissipated P = I2R = (2A)2(1000Ω) = 4000W Voltage across resistor V = IR = (2A)(1000Ω) = 2000V
The rule for voltage across each resistor in a series circuit is that the total voltage supplied by the source is equal to the sum of the voltage drops across each resistor. In a parallel circuit, the voltage across each resistor is the same and equal to the source voltage.
The power dissipated by a 10 ohm resistor with 800v across it is 64 kw.Ohm's law: current is voltage divided by resistancePower law: power is voltage times current, so power is voltage squared divided by resistanceDon'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.
No. If a voltage is applied across a resistor, a current flows through it.
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.
Like Ohm's Law, the formula for calculating power is a simple product of two quantities. It is given by the formula P = VI, where V is the voltage in volts and I is the current in amperes (or simply amps). So, if you know the value of any two of the quantities, you can easily calculate the third with simple arithmetic. For example, if the current flowing through a resistor is two amps and the voltage drop across that resistor is five volts, the power dissipated by the resistor is, P = VI = 5 volts * 2 amps = 10 watts. If you are given the power and the voltage, you can easily find the current. For example, if you are told that the voltage drop across a resistor is five volts and is dissipating 10 watts, the current through the resistor is 10 watts/5 volts = 2 amps.
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 voltage across the resistor is whatever voltage is applied. The only maximum here would be a voltage that would damage the resistor. If you think this might happen, you'll have to look up such a voltage from the data sheets.
It doesn't. In a series circuit, the largest voltage drop occurs across the largest resistor; the smallest voltage drop occurs across the smallest resistor.