V = i*r
v = 2 * 60
v= 120v
200 volts.
low current high voltage power dissipation in power line = I2R the resistance of the power line is hard to reduce, especially when it is a long transmission line. but reducing the current through the line reduces losses as the square, a dramatic savings. reducing voltage would have no effect and would dramatically increase losses due to increase in current to try to deliver same power.
The line current increases when more bulbs are switched on in parallel, since more parallel current paths results in lower effective resistance. The line voltage should not change in response to any normal use of electric power in a single house.
In a Silcon diode no current flows in the forward direction (anode to positive voltage) until approximately 0.6 - 0.7Volts is reached. Above this voltage the current rises in line with Ohms Law. In the reverse direction only micro Amps flow (leakage current) In a Germanium diode the threshold is about 0.2 volts and reverse leakage is higher.
The Current-Voltage relationship of a diode is not constant (not a straight line) and hence the resistance cannot be measured. Due to this non-linear nature of the the curve, there exists a unique value of resistance at every point of the curve which is called dynamic resistance (not static of constant resistance). The dynamic resistance equals the change in voltage divided by the change in current, when the voltage is changed by a small amount. In other words it is the slope of the graph of voltage against current. The dynamic resistance is different at different current values. About 30 years ago, and I do not remeber the brand or maker, there was a digital multimeter that DID measure dynamic resistance in diodes. It was a God Send for testing diodes in circuit. Diodes only conduct in one direction, so the device would show an open in one direction and a resistance under 1000 ohms on the other or a short (0 ohms).
A watt meter that has current through its current coil and voltage across its voltage coil will indicate zero if the power factor between the volts and amps is zero. This condition would be one in which the current will either lead or lag the voltage by 90 degrees and the circuit will have amps, volts, VARS, and VA, but will not have Watts.
Current = Voltage / resistance (more properly impedance) so the current will be 220/55 or 4 amps.
The answer is 20 divided by 40, in amps.
In short, no. There are ways to make the voltage rise, but inverting the power will not help you. If you don't care about the amount of Amps you have, the simplest way to raise the voltage would be to add resistance to the line. Volts and amps are linked by Ohm's law, which states that V=IR, where V is volts, I is current (in amps), and R is resistance in ohms. Therefore you can create a higher voltage at the cost of amperage by adding resistance to your line.
The amps will be the same. Volts will depend on between which two points you're measuring it.
Ohm's Law: Voltage = Current times Resistance
In star the voltage from line to neutral is 1/sqrt(3) times the nominal voltage, while the load current equals the line current. In delta the voltage between lines is the nominal voltage, while the load current is 1/sqrt(3) times the line current (for a balanced load). So a delta load needs 3 times the resistance compared to a star load of the same power.
For practical purposes, transformers do not lose power during transformation. Thus, if you have the VA of the transformer, you can simply divide the VA by the voltage to get a rough idea of maximum current. In a real application, the maximum amperage will be dependent on the phase angle of the transformer.
Yes, 220 and 240 volts are a nominal figure in the same voltage range. It is brought about by the power company, as they have a responsibility to keep voltages within a certain 10% range. The load will only notice a difference of 1% on the load current. E.g. wattage load of 2400. Amps = watts/volts. 2400/240V = 10 amps. 2400/220V = 10.9 amps. On a constant resistance as the voltage goes lower, the current goes higher and vise versa as the voltage goes higher, the current goes lower.
Voltage drop is resultant of IR ie current and the line resistance, not dependent on impressed emf
Watts is a unit power and it is found muliplying Volts x Amperes, so how many amps to 230 watts depends on the voltage. If it is line voltage 120 V then 230 divided by 120 the result is a little less than 2 Amps RomeoTango
It depends on the voltage on line side. KVA is simply thousand volt-amps, so you need to know voltage in order to calculate amperes.Another AnswerThe rated primary current is the rated apparent power of the transformer, divided by the rated primary current. However, the actual primary current is determined by the actualsecondary load current in proportion to the reciprocal of the turns ratio.
Total voltage = the source. The voltage around the circuit is divided proportionally by each of the resistances in line. The current is = the source voltage divided by the sum of all the resistance.