0
If the voltage dropped across a resistor increases by a factor of 10 the power dissipated by the resistor is?
Wiki User
∙ 7y agoWant this question answered?
Be notified when an answer is posted
Add your answer:
Earn +20 pts
What process will double the power dissipated by a resistor?
Increase the voltage across the resistor by 41.4% .
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.
When a 3 ohm resistor draws a current of 4 A when hooked up to a battery of unknown voltage what is the battery's voltage and what is the power dissipated by the resistor?
Voltage can be calculated using Ohm's Law:Voltage = Current (A) x Resistance (Ω)Voltage = 4A x 3Ω = 12 VoltsTherefore, the battery is a 12 Volts.The power dissipated is Voltage x CurrentPower = 4A x 12V = 48 Watts
How much work does a battery connected to a 21 ohm resistor perform in one minute?
That completely depends on the voltage of the battery.The energy delivered by the battery and dissipated by the resistor in one minute will be[ 60 x (Voltage of the battery)2 / 21 ] joules
What power is dissipated if 800v is applied across a 10 ohm resistor?
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.
What process will double the power dissipated by a resistor?
Increase the voltage across the resistor by 41.4% .
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.
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 current is drop when it flowing through the resistor?
Current flows in loops, voltage drops across elements. With relation to current, what flows in, must flow out, so no, current is not dropped across a resistor, it flows through a resistor and voltage is dropped across the resistor.
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).
When a 3 ohm resistor draws a current of 4 A when hooked up to a battery of unknown voltage what is the battery's voltage and what is the power dissipated by the resistor?
Voltage can be calculated using Ohm's Law:Voltage = Current (A) x Resistance (Ω)Voltage = 4A x 3Ω = 12 VoltsTherefore, the battery is a 12 Volts.The power dissipated is Voltage x CurrentPower = 4A x 12V = 48 Watts
Does current flow through a resistor or across a resistor?
Normally through the resistor's internal construction. It flows through any part of the resistor that has low resistance- be it anywere. And then there's this. It might be that one should consider that current flows through a resistor and voltage is dropped across a resistor. Perhaps this is where the question began. The former is fairly straight forward. The latter can be vexing. Voltage is said to be dropped across a resistor when current is flowing through it. The voltage drop may be also considered as the voltage measureable across that resistor or the voltage "felt" by that resistor. It's as if that resistor was in a circuit by itself and hooked up to a battery of that equivalent voltage.
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
How much work does a battery connected to a 21 ohm resistor perform in one minute?
That completely depends on the voltage of the battery.The energy delivered by the battery and dissipated by the resistor in one minute will be[ 60 x (Voltage of the battery)2 / 21 ] joules
What power is dissipated if 800v is applied across a 10 ohm resistor?
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.
What is the formula of power being dissipated in a resistor?
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.
How do you measure current without an amp meter?
Depends on the current. Put a resistor in-line with the current, then measure the voltage across the resistor. V=RI. So, divide the measured voltage by resistor value. Be careful with the size of the resistor, as Power dissipated in a resistor is R*I^2 or V^2/2. So, a 1-Amp current into a 1 Ohm resistor will result in a 1Watt power dissipated in the resistor. If it's too small, it'll burn. Also, notice that if you do that, you haven't measured the current in the original circuit. You've measured the current when an extra resistor is installed in the original circuit, and that's different.
