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How much current is needed to produce 50 mA of current through a 240Ω resistor?
To determine the voltage needed to produce 50 mA (0.05 A) of current through a 240Ω resistor, you can use Ohm's Law, which states ( V = I \times R ). Plugging in the values, ( V = 0.05 , \text{A} \times 240 , \Omega = 12 , \text{V} ). Therefore, a voltage of 12 volts is required to produce a current of 50 mA through a 240Ω resistor.
What voltage is developed across a 330 Ω resistor if 100 mA of current flows through it?
33V
How much resistance is required to limit the current to 1.5 mA if the potential drop across the resistor is 6V?
Using ohm's law, V=IR then R=V/I =6/0.0015=4000 ohm = 4k ohm resistor.
How much current flows through a 5k resistor connected to a 20 volt source?
Use the equation V = I * R 20 = I * 5000 I = 20 / 5000 I = 0.004 Amps The answer to your question is 4 mA of current will flow through the resistor.
The voltage drop measured across a 2K ohm resistor is 24 mV How much current is flowing through it?
0.012 mA
If the current flowing trough a 10 ohm resistor is 15 mA the power consumed in the resistor is?
The power consumed in a resistor can be calculated using the formula ( P = I^2 R ), where ( P ) is power, ( I ) is current, and ( R ) is resistance. In this case, with a current of 15 mA (or 0.015 A) flowing through a 10 ohm resistor, the power consumed is ( P = (0.015)^2 \times 10 = 0.00225 \times 10 = 0.0225 ) watts or 22.5 mW.
What happens to the current flowing through a metal resistor when the voltage across is increased?
When the voltage is increased across a metal film resistor, the current flow will also increase. Ohm's law states that the current flowing through a resistor is directly proportional to the voltage across that resistor. I = V/R Let us assume an initial voltage drop across a 4.99K ohm metal film resistor is 5V. The current flow through the resistor is calculated to be: I = 5/4990 = 0.001 Amps or 1 mA If that voltage were to say double to 10V: I = 10/4990 = 0.002 Amps or 2 mA Using these values it is also possible to calculate the power dissipated by the resistor. P = I*V = 0.002 * 10 = 0.02 Watts This power calculation determines the minimum physical case size needed for the resistor to function within these conditions. Anything smaller, the resistor will fail.
What is the power consumed in a 10 k ohm resistor due to a periodic synusoidal current of maximum magnitude 90 mA passing through it?
P=IE What voltage (E) will be dropped across the resistor? Current (I) = .090 A Assuming 90 VAC dropped across the resistor, then P=90 x .090 which = 8.1 Watts.
How much voltage is required to cause 8mill of current to flow through a 10kilo resistor?
V = IR = 8x10-3 (8 mA) * 10x103 (10k) = 80v
What is the voltage drop across a 100 W resistor when the electric current flowing through it is measured at 250 mA?
Who can tell? The power rating of a resistor simply tells us the maximum power that resistor is capable of handling; it doesn't tell us anything about the actual power being produced for any given current. So, to find out the voltage drop across that resistor, you will need to find out its resistance, and multiply this value by the current you specify.
What is the current through a 400 ohm resistor with 6 volts across it?
Use Ohm's Law - in this case, solving for current: I = V/R (current = voltage divided by resistance). Since you are using standard SI units, the answer will be in amperes.
What is the power when there are 58 m'A of current through a 4.7 kilo-ohm resistor?
By Ohm's law, voltage = current times resistance, so 50 mA of current through a 4.7 kOhm resistor yields a voltage drop of 235 volts.By the power law, power = voltage times current, so 50 mA of current and 235 volts yields a power of 11.75 watts.That is quite a bit of power, so the resistor is going to become very hot, and must be rated for that power, otherwise it will self-destruct and/or cause a fire.AnswerAn alternative, and quicker, method of solving this problem would be to simply to use the following equation:P = I2RIn which case:P = (58 x 10-3)2 x (4.7 x 103) = 15.81 W (Answer)(***The original answer used 50 mA, rather than 58 mA, by mistake.)
