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What is the current in a circuit that has a resistance of 30.0 and a power of 2.00 W?
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∙ 9y ago
0.258 A
Rachoon Man
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How do you calculate short circuit current?
A short circuit = 0 ohm, from this one can use ohms law to calculate the current, that is E/R voltage divided by resistance as in this instance your resistance is zero and you can't divide by zero so your current will be infinit. In other words if your power supply is large enough to supply 200 amp, your current on a short circuit will be 200 amp.
What happens if a 200 mA power supply is used on a circuit needing 2 amp?
If a 200 mA power supply is used on a circuit needing 2 amp it would overload the power supply. Two amps equates to 2000 milliamps. Trying to pull 2000 milliamps from a device that is only capable of supplying 200 milliamps is not advisable. The excessive current draw would most likely burn the wires open and render the power supply useless.
What voltage is necessary to produce a current of 200 amperes through a resistance of 1 X 10-3 ohms?
voltage = current x resistance, so: voltage = 200 x 0.001 = 0.2 volts
Does adding resistors in series increase or decrease the overall resistance of a circuit?
Resistance in series adds together. two 100 ohm resistors in series are equivalent to one 200 ohm resistor. to make an equation out of it ( even though it is simple) you can say: Rtotal=R1+R2+R3...+Rn
Can I swap the IN5408 with BY255 diodes in power stage of my tube amp?
The 1N5408 is rated 3 amperes forward current and 1000 volt reverse voltage, with a 200 ampere peak surge current rating. The BY255 is rated 3 amperes forward current and 1300 volt reverse voltage, with a 100 amperes peak surge current rating. The two diodes are comparable, but the 1N5408 has a higher transient rating. Since it does not seem that you are talking about a power supply application, you can probably substitute the BY255 for a 1N5408. It depends on the circuit design, but most circuits (if properly designed) are over-rated so, again, it would probably work. I would not use the BY255 in the power supply, however, due to the short term transient that comes from turning on the power. Again, circuit design prevails, and it might work even there, because inrush current will be limited by the inductance of the transformer and the dynamic resistance of the diode.
Diode reverse breakdown14v current of 70mA and an un-stabilised supply voltage of 28v calculate resistance?
If a circuit element has a voltage of 14V and a current of 70mA, then the resistance of the circuit element is 200 ohms. This is ohm's law. The resistance or type of the power supply is meaningless.
How do you calculate short circuit current?
A short circuit = 0 ohm, from this one can use ohms law to calculate the current, that is E/R voltage divided by resistance as in this instance your resistance is zero and you can't divide by zero so your current will be infinit. In other words if your power supply is large enough to supply 200 amp, your current on a short circuit will be 200 amp.
200 volts consumes how many watts of power?
Actually voltage doesn't CONSUME power. But to answer what I think is your question, it is ENTIRELY dependent on the resistance (R) of the circuit, which in turn denotes the current flow (I). you must know one or the other to calculate power ( P=EI)
Four 50 ohms are connected in series calculate the current in the circuit?
The series resistance is 4 x 50 = 200 Ohms. You would need additional information to get the current; usually this is calculated from the voltage. current = voltage / resistance.The series resistance is 4 x 50 = 200 Ohms. You would need additional information to get the current; usually this is calculated from the voltage. current = voltage / resistance.The series resistance is 4 x 50 = 200 Ohms. You would need additional information to get the current; usually this is calculated from the voltage. current = voltage / resistance.The series resistance is 4 x 50 = 200 Ohms. You would need additional information to get the current; usually this is calculated from the voltage. current = voltage / resistance.
The total resistance for this circuit if all resistors are 200 ohms is?
400 ohms
How any volts are produced in a circuit with 40 amp's of current flowing through a wire with 5 ohms of resistance?
It is a voltage (potential) applied to a load that causes a current to flow through the load. Ohm's Law encapsulates this principal and states Volts = Current x Resistance. In your example, the applied voltage would be 200 volts.
How much curretn is in a circuit that includes a 9- volt battery and bulb with a resistance of 12 ohms?
Current, I is equal to V (voltage) divided by R (resistance); Hence: I=V/R = 9V/200 ohms: I = 0.045 Amps, or 45ma (milliamps).
A light draws current of 1.1 amps at 230 volts the power onput is 200w what is the circuit power factor?
The current's power factor is the true power divided by the apparent power. The Apparent Power is the volts multiplied by the amps. In this example, the ratio would be 200/253, or approximately .79.
Will the current in a light bulb connected to a 200-v source be greater or less than when the same bulb is connected to a 110-v circuit?
It will be greater:current = voltage / resistanceSince the bulb's resistance doesn't change, then current is a direct function of voltage.
A desktop computer and monitor together draw about 0.6 A of current They plug into a wall outlet that is 120 V What is the resistance in ohms of the computermonitor combination?
To find the resistance in a circuit, use the equation R = V/I, where R is the resistence in ohms, V is the voltage, and I is the current in amps. Therefore, your equation is R = 120/0.6. Therefore, the resistance is 200 ohms. Hope this helps
What is the voltage when current is 0.2 amperes?
What some people write as 0.2 Amps is the same as 200 milliAmps or 2x1/10Amp.
How do you increase voltage?
You can put less resistance (more load) on the battery with larger wires, but if you exceed a particular current output for a given duration, you will overheat the battery. To safely increase current output, use two batteries connected in parallel.
