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In a circuit there is a 0.8 a circuit in a bulb the voltage across is 4.0 what is the bulb resistance?
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When you add another light bulb to a circuit does it effect the voltage?
When a bulb is added in parallel to a circuit with a single bulb, the resistance of the circuit decreases. This is because the existing bulb's operating point remains the same, due to the fact that Kirchoff's voltage law states that the signed voltages around a series circuit must add up to zero, with the two bulbs being their own series circuit. A consequence of this is that the voltage across both bulbs must be the same, i.e. that the voltage across the first bulb does not change. The second result of adding the bulb is that the current in the overall circuit increases. This is because the second bulb must pull some current in order to operate, yet we know that the voltage across the first bulb did not change. As a result, due to Ohm's law, the current through the first bulb did not change. And, finally, since Kirchoff's current law states that the signed sum of the currents entering a node is zero, the addition of a second load in parallel with the first load must, therefore, represent additional current.
An increase in resistance in a circuit will cause?
In a simple circuit, lowering the voltage will not cause the resistance to do anything. Lowering the voltage will, however, cause the current to also lower.This ignores temperature coefficient. If there is substantial power involved, a typical bulb, for instance, will grow cooler and its resistance will decrease when you lower the voltage, but that is usually a small effect.
Watt is the current allowed to flow through a 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.
What voltage is required to sustain a current of 0.50 through a light bulb resistance of 190?
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
How does the brightness of each bulb in a parallel circuit compare to the brightness of the bulb in a simple circuit?
The brightness of each bulb in a parallel circuit is the same as the brightness of a bulb in a simple circuit. By Kirchoff's voltage law, each element of a parallel circuit has the same voltage drop across it. With the same voltage, the same type of bulb will dissipate the same power, and have the same brightness.
What happens when one bulb is added to a series circuit?
The resistance is increased, the voltage across each bulb is decreased and the current through the circuit is reduced.
Suppose in a circuit with 2 bulbs the resistance of bulb 1 is greater than that of bulb 2. how will the voltage across the two bulbs compare?
if the resistance of bulb A is 2x that of B then there will be twice as much voltage across it (ratio 2:1 ). both voltages shall equal the system voltage assuming they are in series and there are no other components in the circuit if the bulbs are in parallel the voltage across them will be equal and that of the system
Why is the voltage across a battery in a parallel circuit equal to the voltage across each bulb?
The voltage across a battery in a parallel circuit is equal to the voltage across each bulb because Kirchoff's Voltage Law (KVL) states that the signed sum of the voltages going around a series circuit adds up to zero. Each section of the parallel circuit, i.e. the battery and one bulb, constitutes a series circuit. By KVL, the voltage across the battery must be equal and opposite to the voltage across the bulb. Another way of thinking about this is to consider that the conductors joining the battery and bulbs effectively have zero ohms resistance. By Ohm's law, this means the voltage across the conductor is zero, which means the voltage across the bulb must be equal to the voltage across the battery and, of course, the same applies for all of the bulbs.
What is the difference between voltage output of a battery and the voltage across each DC bulb in a series circuit?
Exactly...you answered your own question. Each DC bulb will drop voltage according to its resistance and the amount of current it draws.
What is the factor that affects the brightness of the bulb?
number and voltage of the cells in the circuit resistance of each bulb
When you add another light bulb to a circuit does it effect the voltage?
When a bulb is added in parallel to a circuit with a single bulb, the resistance of the circuit decreases. This is because the existing bulb's operating point remains the same, due to the fact that Kirchoff's voltage law states that the signed voltages around a series circuit must add up to zero, with the two bulbs being their own series circuit. A consequence of this is that the voltage across both bulbs must be the same, i.e. that the voltage across the first bulb does not change. The second result of adding the bulb is that the current in the overall circuit increases. This is because the second bulb must pull some current in order to operate, yet we know that the voltage across the first bulb did not change. As a result, due to Ohm's law, the current through the first bulb did not change. And, finally, since Kirchoff's current law states that the signed sum of the currents entering a node is zero, the addition of a second load in parallel with the first load must, therefore, represent additional current.
Suppose in a circuit with two bulbsthe resistance of bulb 1 is greater than that of bulb 2how will the voltage across the two bulbs compare?
That depends on whether the bulbs are wired in series or in parallel.
Why does a light bulb light in a closed circuit?
A bulb does not light up if there is no voltage available across the bulb, or if the bulb is burned out.
Why does the voltage drop as electricity flows through a light bulb?
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.
An increase in resistance in a circuit will cause?
In a simple circuit, lowering the voltage will not cause the resistance to do anything. Lowering the voltage will, however, cause the current to also lower.This ignores temperature coefficient. If there is substantial power involved, a typical bulb, for instance, will grow cooler and its resistance will decrease when you lower the voltage, but that is usually a small effect.
Why will the bulb not light up in a short circuit?
That's because the path of the short circuit provides a much lower resistance than the actual path and which the current will choose to flow through the path with a lower resistance rather than the path connected to a bulb,which explains that why the bulb won't light up.
Would you show me a diagram of a series circuit?
Diagrams are not supported with WikiAnswers, sorry, so you will have to use your imagination...A simple series circuit can be built with a battery, a switch, and a light bulb. One end of the battery is connected to one end of the switch. The other end of the switch is connected to one end of the bulb. The other end of the bulb is connected to the other end of the battery.If the switch is open, no current flows, and the bulb does not illuminate. If the switch is closed, current flows, and the bulb illuminates. By Ohm's law, the current through the bulb is proportional to the battery voltage and inversely proportional to the resistance of the bulb. Note, of course, that we are talking about hot resistance of the bulb, because cold resistance is an entirely different thing, due to the temperature coefficient of the bulb. Also, by Kirchoff's current law, the current at every point in this simple series circuit is the same and, by Kirchoff's voltage law, the voltage across the battery is the same as the voltage across the bulb.
