<p>Hi!</p>
<p>I cant understand this exercise:</p>
<p>Three identical light bulbs are connected to a source of emf. What will happen to the light intensity of the other 2 if one of the 3 bulbs burns out?</p>
<p>In my opinion the intensity of the other 2 would increase because the total resistance would decrease, therefore the current would increase and the power (P=U.I) would increase too.</p>
<p>But in the book says that the light intensity wouldn’t change.</p>
<p>Can anybody help me?</p>
<p>wat duhh…</p>
<p>I’d think that it would disrupt the circuit…since a connection burned out, so all the bulbs would turn off…
i think</p>
<p>But if ^ thats not the case, then I’d have to agree with you.</p>
<p>i think if it was a parallel circuit the light intensity would stay the same but if in was in a series the light intensity would decrease.</p>
<p>yeah, if it’s parallel the intensity does not change, but in series none will light at all.</p>
<p>^why would none of them light in a series? bit rusty on my circuits…</p>
<p>Had they been in series, the bulbs’ light intensity would decrease since the potential difference would have to be divided by two resistors , but I guess it was parallel so the same potential difference applied on both…</p>
<p>^yeah that’s what I thought</p>
<p>No… if they were in series and one bulb burned out the circuit would stop.</p>
<p>I guess so…</p>
<p>yeah the lightbulb itself is a piece of the circuit and if it blows out the circuit is broken (in series). Therefore, that is why none would light.</p>
<p>Each bulb behaves as a resistor (in this case they not only dissipate energy as heat but also as black body photon radiation).</p>
<p>In parallel:
Assuming an ideal power supply (constant voltage and current source, at any load) the remaining bulb will remain at the same brightness. In real world conditions, the load of the extra bulbs would have caused a voltage drop in the supply (along with a brightness drop in all three bulbs) and as a result the final bulb will become somewhat brighter (depending on previous supply voltage drop).</p>
<p>In series the remaining bulb won’t have a complete circuit and will therefore go dark. The only exception could be very high frequency and voltage AC which could still light the bulb through capacitive charge discharge (like flyback streamer in a plasma globe). They won’t be dealing with odd scenarios like this on the exam though so just assume an ideal supply of DC unless otherwise stated.</p>
<p>You can also use your power formula for resistors (or bubs, will are not very good approximations to black bodies BTW) to see what happens in the parallel case: P= V^2 / R. For parallel bulbs, no matter how many there are, V is the same as the EMF voltage, so the power used by each bulb (and their brightness) doesn’t change.</p>
<p>For series bulbs, here’s the kind of question I would expect: one of the bulbs short-circuits (i.e., becomes a piece of wire with R=0). What happens to the brightness of the other bulbs now?</p>
<p>Just took the blue book physics test (got 800 BTW) and saw that question. It had one bulb in series with an SPST switch and all three in parallel with the source. Oddly, they seem to be using a “real world” non-ideal power supply in this problem, making the correct answer E, all three bulbs change in brightness. Obviously the bulb with the switch goes out. But the fact that they assume the power supply to be non-ideal (it is a battery, but still this is a mainly theoretical test) means that the other two bulbs now see more voltage and current, causing them to become brighter.</p>
<p>True, bulbs are not ideal blackbodies, but they are a rough approximation. This will not be on the test :-D</p>
<p>If I saw this question on the real test, I might complain that they don’t specify if its an ideal or non-ideal power source.</p>