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.
To determine the current flowing through each bulb, you would need to know the voltage (V) applied across the bulbs. According to Ohm's Law (I = V/R), where I is the current in amperes, V is the voltage in volts, and R is the resistance in ohms, you can calculate the current. For example, if the voltage is 120 volts, the current would be I = 120V / 100Ω = 1.2 A. Without the voltage value, the current cannot be calculated.
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).
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 current passing through the bulb is directly related to the voltage applied across it and the resistance of the bulb. Using Ohm's Law (I = V/R), we can calculate the current flowing through the bulb by knowing the voltage and resistance values. Additionally, the brightness of the bulb can also be an indicator of the current passing through it, as higher current typically results in a brighter bulb.
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.
To determine the voltage required to pass 0.5 A of current through a 4.5-W bulb, you can use the formula P = V * I, where P is power, V is voltage, and I is current. Given that power is 4.5 W and current is 0.5 A, you can rearrange the formula to solve for voltage: V = P / I. Plugging in the values, V = 4.5 W / 0.5 A, which equals 9 volts. Therefore, 9 volts is required to pass 0.5 A of current through a 4.5-W bulb.
The current flowing through the 75-watt light bulb connected to a 110-volt wall outlet can be calculated by using the formula: Power = Voltage x Current. Therefore, the current flowing through the light bulb would be approximately 0.68 amps.
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.
The current in the light bulb will be greater when connected to the 200-v source compared to the 110-v circuit, assuming the resistance of the light bulb remains constant. This is because current is directly proportional to voltage in an electrical circuit according to Ohm's Law (I = V/R), so a higher voltage will result in a greater current flow through the bulb.
it remains same i=voltage/total 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