Wiki User
∙ 13y agoPower = E2 / R
Resistance = E2 / power = (240)2 / 3,000 = 19.2 ohms
Current = E / R = 240 / 19.2 = 12.5 amperes
Wiki User
∙ 13y agoYou have to imagine the internal resistance as being in parallel with any load you connect. You get the maximum possible current when the load is zero. In this case, just apply Ohm's Law. That is, divide the voltage by the internal resistance.
Assume the rating of 100W refers to operation on a supply of 117 volts.Power = (voltage) x (current)Current = (power) / (voltage) = 100/117 = 0.855 ampere (rounded)Power = (voltage)2 / (resistance)Resistance = (voltage)2 / (power) = (117)2 / 100 = 136.89 ohms
Power in a circuit is inversely proportional to the resistance, all other things being equal. Voltage equals amperes time resistances, so amperes equals voltage divided by resistance. Watts equals voltage times amperes, so watts equals voltage squared divided by resistance.
Load!
You need to divide the supply voltage by the impedance of the load. The impedance of the load is the vectorial sum of its resistance and reactance, where reactance is proportional to frequency.
Yes, increasing the voltage to an electric heater will typically result in an increase in the amperage it draws. This is due to Ohm's Law, which states that current (amperage) is directly proportional to voltage and inversely proportional to resistance. As voltage increases, the current drawn by the heater will also increase.
If the terminal voltage decreases when more current is drawn, that is due to the internal resistance of the power supply. Every power supply has a limit to how much current can be drawn. It is limited by the internal resistance and due to ohms law the more current drawn through a resistor, then the more voltage is produced across it. This is in opposition to the terminal voltage and is subtracted from it.
this happen due to sudden amount of voltage drop in the main feeder due to large current drawn by the heater , so this drop in voltage will let the bulb operate by avltage of a mount (V-Vd) [where V represent the supply voltage and Vd represent the voltage drop in the main feeder of the circuit] , which is less than the voltage before the heater is connected and due to this situation the current passes through the wire of the bulb will be less and therefore the brightness of the bulb becomes dim.after a while the wire of the heater will has high temperature which increase its resistance and due to this the current drawn by the heater will decrease than the current at the first time , therefore the drop in voltage will also decrease , which implise increase in current drawn by the bulb and therefore the dimness decrease .
It causes the battery's voltage to drop when a current is drawn from it.
It depends. If voltage is drawn along the horizontal axis, then the slope at any point on the graph represents the reciprocal of resistance at that point. If current is drawn along the horizontal axis, then the slope at any point on the graph represents the resistance at that point.
Because by increasing the load resistance, the total circuit resistance is reduced. This means with less resistance, there is more current drawn from the source. Doubling the size of a load resistor increases the load current.
Is this, intentionally, a trick question?We are dealing with alternating current, here, not direct current. So, if you divide the supply voltage by the current drawn by the television set, you are determining its impedance(Z), not its resistance:Z = V/I = 120/3 = 40 ohmsImpedance is the vector sum of resistance and reactance. As the current is probably being drawn by a transformer, the resistance will be significantly lower than the reactance, perhaps only an ohm or two -if that!So, from the information supplied, you cannot determine the resistance.
The maximum current that a cell can deliver flows when the resistance between the terminals of the cell is zero. This situation occurs when the terminals are connected by a conductor with very low resistance, such as a thick wire or a wrench. But not for long.
No, lower voltage equipment will not operate on a higher voltage because the wattage or current drawn by the equipment will be higher that the rating of the equipment at the lower voltage. For example if you take a heater rated at 5000 watts at 208 volts, the current is I = W/E 5000/208 = 24 amps. The resistance of this heater is R = W/I (squared) = 5000/24 x 24 (576) = 8.6 ohms. Applying 240 volts to this heater whose resistance is 8.6 ohms results in this new heater wattage rating. W = E (squared)/R = 240 x 240 (57600)/8.6 = 6697 watts. As you can see the original wattage of the heater is overloaded by about 25% of what the manufacturers specified the wattage to be. Ohms law states that the current is directly proportional to the applied voltage and inversely proportional to the resistance of the circuit. W = watts, I = amperage, R = resistance in ohms and E = voltage.
You have to imagine the internal resistance as being in parallel with any load you connect. You get the maximum possible current when the load is zero. In this case, just apply Ohm's Law. That is, divide the voltage by the internal resistance.
A loose battery terminal will cause intermittent power supply, high resistance and heating. A high resistance will cause a voltage drop, as more current is drawn from it.
parallel means they will never cross each other. take a example when a resistance connected in parallel then a current flowing through it(both resistance) is not same depended on the value of each resistance. while in the series circuit current values is same for both resistance. parallel lines are drawn below:- series lines:- -------