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I understand that increasing voltage increases the energy given to each coulomb in the circuit, so in theory more energy given to a bulb should make a bulb glow brighter. But also if voltage is constant and resistance decreases, then current increases so that means that the bulb glows brighter too because more coulombs are flowing through it per second. But what happens if voltage and resistance are increased proportionately so that current remains the same? More energy is supplied to each coulomb so it should make the bulb grow brighter right? Or have I made some error?

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    $\begingroup$ The bulb with get brighter. Power is equal to $V^2/R$, so if you increase V and R at the same rate then power will increase. Note that with real incandescent light bulbs the resistance of the filament increases a lot from the bulb's "off" state to its "on" stage due to the fact that the resistance depends on temperature. $\endgroup$ – Samuel Weir Nov 22 '15 at 5:22
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Without changing the characteristics of the light bulb, it's not possible to change the voltage if you want the current to stay the same. When the voltage is increased, the current through the bulb has to increase as well, because of Ohm's law: $I=V/R$ which means that if $V$ increases, so does $I$.

If you want the current to stay the same when the voltage increases, you need to insert additional resistance in series with the bulb's filament. That additional resistor will show a voltage drop of $V_1=I*R_1$, which is exactly the increase in applied voltage, and the voltage over the bulb will still be the same as the original. This extra resistor now dissipates the additional power.

If this extra resistor is part of the bulb's filament (so we now have a different bulb), then that extra power will make the bulb brighter, because $P=V*I$.

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If you are comparing two voltages with identical currents, you cannot be talking about the same bulb in both cases. This means that you are comparing two different bulbs, and there is no way to tell which will be brighter, since different bulbs can be designed for different luminous efficacy, which is light per unit power. For instance, a bulb can be designed with a longer filament and operated at a lower temperature and produce less light, but last longer since the filament is run cooler.

If you are comparing two bulbs with similar efficacy, then the bulb run at the higher voltage will be brighter due to its greater power dissipation.

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  • $\begingroup$ Thanks for your answer but I already thought of an answer: P=VI. At the same current, power out will be greater, so the bulb glows brighter. $\endgroup$ – Airdish Nov 22 '15 at 5:18
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    $\begingroup$ Sorry, no (or, not necessarily). Power may be greater, but that does not meant the filament is hotter and putting out more light. If you're talking about the same bulb in both cases, equal current implies equal voltage, so power is the same. Since you're saying different voltage for the same power this means different bulbs. Different filaments will not produce the same light for the same power, so you just can't tell which lamp will be brighter. $\endgroup$ – WhatRoughBeast Nov 22 '15 at 6:07
  • $\begingroup$ @Airdish It may be more intuitive to write P=V I(V) . Current is a function of voltage for the device. That function is roughly I(V)=kV. So P=kV^2 . That's true for all memoryless (resistive) devices. You could come up with a device that let's you control it's current for any input voltage (say a variable resistor). In that case all you can say is the consumed power is P= VI, but you cannot take further conclusions as luminosity or such. $\endgroup$ – Real Jul 30 '16 at 4:24
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Yes, It will get brighter up to a point of failure, but no standard bulb can do that. Your question framed the conditions as 'If voltage and resistance are increased proportionately keeping current constant'. The intensity of light produced from a given bulb comes the resistance built into the bulb. Each bulb has a maximum voltage potential that it can utilize or withstand from that design, but its resistance is fixed.

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When voltage is increased , according to $P=VI$ power will increase. Consequently the bulb will glow brighter.

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Distance if the distance stays the same then yeah but if not than no. and a different type of connector Adding a resistor The current depend from where are you measuring it, it will be some different throughout the circuit.

As well An electric current is a flow of electric charge. In electric circuits this charge is often carried by moving electrons in a wire. It can also be carried by ions in an electrolyte, or by both ions and electrons such as in a plasma.

Nop Nop the current through what!! the wir ? light? It really different on the type of the light some are programmed to stay the same or don't you lot of voltage they putt gas so it will not use as much energy as it produce light and heat. So Answer is the light will blow up! if the voltage is high and it depends how much you're adding.

No the next Einstein. enter image description here

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protected by AccidentalFourierTransform Aug 11 '18 at 13:47

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