Ohm's Law: Resistance is voltage divided by current
Power Law: Power is current times voltage
Combining them gives: Resistance is voltage squared divided by power
220 volts squared divided by 100 watts = 484 ohms. Note that this is hot resistance. If you measure the bulb in the cold state, you will get an entirely different, smaller, value, due to the extreme temperature coefficient of the filament.
Independently of that, since you ask for peak voltage, that means you are talking about an AC voltage source. We have to assume a sinusoidal waveform, and that the 220 volts was the RMS value. In this case, the peak value is simply the RMS value multiplied by the square root of 2, i.e. 0.707..., making the peak value 311 volts.
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
The voltage of a circuit with a resistance of 250 ohms and a current of 0.95 amps is 237.5 volts. Ohms's law: Voltage = Current times Resistance
No it will not. If you need increase the supply voltage and remove the choke.
If a rheostat is connected in parallel with a light bulb, the setting of the rheostat should have no effect on the performance of the light bulb, as long as the power supply is able to maintain its output voltage and deliver the current demanded by their parallel combination.
The higher the resistance the dimmer the light will become. The voltage drop (current) is proportional to the resistance as seen in the equation V=IR, since voltage remains the same throughout a series circuit if the equivalent resistance goes up the amount of current reaching the lightbulb must go down to equal the voltage thus creating a dimmer lightbulb.
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
An incandescent bulb has a filament that has a resistance. The value of the resistance determines the current that will flow for a given supply voltage. The heat generated by the current flowing through the filament gives off light. As the resistance of the filament decreases the current increases and you get more light.
A light doesn't output current, it "draws" current based on voltage and its resistance. Voltage = Current x Resistance or Current = Voltage / Resistance. (Ohm's Law)
there is no voltage and resistance
there is no voltage and resistance
To calculate current passing through a light globe, you can use Ohm's Law: current (I) = voltage (V) / resistance (R). To calculate voltage across a light globe, you can rearrange Ohm's Law to solve for voltage: voltage (V) = current (I) * resistance (R). Just make sure you know the resistance of the light globe in ohms.
The voltage of a circuit with a resistance of 250 ohms and a current of 0.95 amps is 237.5 volts. Ohms's law: Voltage = Current times Resistance
You can use Ohm's Law to calculate the current of a light bulb by dividing the voltage across the light bulb by its resistance, which is typically provided on the bulb itself or its packaging. The formula is: Current (I) = Voltage (V) / Resistance (R).
No it will not. If you need increase the supply voltage and remove the choke.
If a rheostat is connected in parallel with a light bulb, the setting of the rheostat should have no effect on the performance of the light bulb, as long as the power supply is able to maintain its output voltage and deliver the current demanded by their parallel combination.
Do you mean why is the voltage in a circuit lower after the light bulb than before it? If so, it's because the light bulb filament has electrical resistance. When an electrical current flows through a resistance, there is a voltage drop across the resistance (Ohm's law).More fundamentally, the light bulb is producing light, which is a form of energy. The voltage drop across the light bulb comes from the fact that electrical energy is being turned into light. If voltage didn't drop, you would be producing energy from nothing. Furthermore, if there were no voltage drop, your circuit would behave the same whether you had no light bulbs, one light bulb, or eighteen million light bulbs - something that clearly can't be the case.
Connecting light bulb is equivalent to connecting a resistance. If you have connected light bulb, there will be some voltage drop across it and your TV may not get sufficient voltage. However if resistance is not big enough, than it wont have any effect.