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The current flowing through a bulb is equal to the (voltage across the bulb) divided by the (bulb resistance), and can be expressed in Amperes. The rate at which the bulb dissipates energy is equal to (voltage across the bulb) times (current through the bulb), and can be expressed in watts.
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
Power is current times voltage, so a current of 0.5 amperes and a voltage of 220v across a bulb will yield a power of 110 watts.
Yes. If voltage leads the current, the impedance is inductive (this would be the case if the load is a motor). If current leads the voltage, the impedance is capacitive (this would be the case for a CFL light bulb).
You know if current is flowing in a bulb circuit because, if there is enough power (voltage times current), the bulb will illuminate. If there is current, but not enough power to illuminate the bulb, you will need to measure the current with an ammeter to see if there is any current.
The current flowing through a bulb is equal to the (voltage across the bulb) divided by the (bulb resistance), and can be expressed in Amperes. The rate at which the bulb dissipates energy is equal to (voltage across the bulb) times (current through the bulb), and can be expressed in watts.
230v power supply .................it requires 12watts......
yes, it can be used - single phase voltage of 230v. 50HZ is important
An incadescent bulb lights up because the voltage source creates a current through the filament of the bulb and it heats up because of its composition and gives off light as a result. No electricity, no voltage and no current.
If you're using it in your home with a voltage supply of 230V [Basically speaking], the resistance would be around 5290Ω. The following equations can help you: P=V*I {Where P is the power, V is the voltage and I is the current} V=I*R {Where R is the resistance}
it remains same i=voltage/total resistance
If you divide the watts of the bulb by the supply voltage, that is the current. For example a 60 w bulb on a 240 v supply gives a current of 60/240 which is ¼ amp.
9 volts====================The question is a bit convoluted.The power dissipated by the bulb and the current through itboth depend on the voltage applied across it.In the real world, the way to ask this question would have to be:If a light bulb dissipates 4.5 W of power when 0.5 A of currentpasses through it, what voltage has been applied across it ?(And, for extra credit, what is the bulb's effective resistance ?)
9 volts====================The question is a bit convoluted.The power dissipated by the bulb and the current through itboth depend on the voltage applied across it.In the real world, the way to ask this question would have to be:If a light bulb dissipates 4.5 W of power when 0.5 A of currentpasses through it, what voltage has been applied across it ?(And, for extra credit, what is the bulb's effective resistance ?)
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
The resistance is increased, the voltage across each bulb is decreased and the current through the circuit is reduced.
Current = (voltage) / (resistance) = 100/130 = 0.769 A = 769 milliamperes (rounded)