The voltage is the main power of a toaster!! but if the current flow is low the heat of the toaster is less heat that you expect, the 8 A is right for your toaster!!!
The resistance of the heater elements is 15 Ohms and the toaster consumes 960 watts.
By Ohm's Law Voltage = Current x Resistance R = V / I = 120 / 12 = 10 Ohms
Assuming DC and resistive loads, resistance equals voltage across the load, divided by the current through it. In this case 120/10 or 12 ohms.
It's the RMS value. A 120 volt lamp (light bulb) is rated according to its RMS voltage. Just like appliances in the home are rated at 120 volts (like your fridge, microwave and toaster), or 220 volts (like your clothes dryer). Note that these appliances will have to stand up to the peak voltage on the AC line. Naturally. And the peak voltage on an AC line is 1.414 times the RMS value of voltage. That means in a 120 volt AC line (120 volts RMS), the peak value of the voltage will be 1.414 times the 120 volts, or right at about 170 voltspeak for each cycle.
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 dc voltage of a rectified ac voltage will be the peak value of the ac voltage less the forward voltage drop of the diode.The rms voltage of a sinusoidal ac voltage is sqrt(peak) / 2, but you also have to consider if the ac voltage is balanced around zero.For a normal US house voltage of 117VAC, the peak voltage is about 165V, or 330V peak to peak. Your dc voltage is then around 164V.Run that rectified voltage through a capacitor, and you will still have 164V peak value, but the voltage over time will dip because the capacitor will discharge during diode off time, and recharge when it turns back on.AnswerA given value of a.c. rms voltage is exactly equivalent to the corresponding value of d.c. voltage. For example, 120 V (rms) is exactly equivalent to 120 V d.c. This is why the alternative name for 'rms voltage' is 'effective voltage'. This is based on the fact that a current of, say, 10 A (rms) will do exactly the same amount of work as a d.c. current of 10 A. And, of course, voltage and current are proportional.
Multiplying 15 amps x 120 volts gives you 1800 watts, or the power consumed by the toaster
Power = (voltage) x (current) = (120) x (0.042) = 5.04 watts
By Ohm's Law Voltage = Current x Resistance R = V / I = 120 / 12 = 10 Ohms
Ohm's Law says Voltage = Current x Resistance V = 5 x 24 = 120 V.
The formula you are looking for is Ohm's Law. Voltage = Current x Resistance (v = I x R). To solve for Current the formula is I = V/R.
The formula you are looking for is W = A x V. Watts = Amps x Volts.
Assuming DC and resistive loads, resistance equals voltage across the load, divided by the current through it. In this case 120/10 or 12 ohms.
Basically, Power = Current*Voltage Current = Power/Voltage Current = 15/120 Current = 0.125A or 125mA
The formula you are looking for is W = A x V. Watts = Amps x Volts.
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To find the resistance necessary, one would need to know how much current the bulb draws. If one knows the current the bulb draws, then one would subtract the 14 volts from 120 volts then divide that by the current the bulb draws and one will find the resistance needed. Once this has been done, one would need to multiply the current drawn by the voltage drop to get the wattage rating necessary. Another important detail to note is that the power dissipated by the resistor will be much greater than the power consumed by the bulb itself. Finally if the bulb burns out the voltage across the contacts will be 120V. I would not recommend using this method to drop the voltage for the bulb.
Yes. If you hook up your phone charger, it is using a little bit over 50% power without a phone attached. No. The voltage potential of 120 volts is at the receptacle outlet slots but to make the current flow, a resistive load is needed. In the case above the phone charger is the load. Un-plug the load and no current flow.