# Why do lightbulbs in parallel stay the same brightness when one is removed

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Let us suppose in the above circuit all the bulbs are identical having resistance r.

In that situation A will be brightest of all the bulbs and next comes B and C as well as D will be lowest bright .

so naturally when you remove either C or D the bulb will be brighter compared to what it was glowing earlier.

I think one should apply the rule that the voltage V = current x Resistance

as the bulbs have identical resistance the bulb A gets the full voltage of the battery. so its glowing brightest getting the max. current say IA.

V = IA.r where V is the battery voltage.

the B, C, and D are sharing the voltage in the lower arm.

V = IB .r + IC.r = IB .r + ID.r

IC and ID are equal as the current IB is dividing equally in C and D arm.

However IC +ID = IB or IC = IB/2. (Kirchhoff's Law)

that's why I noted above that A will be brightest getting current IA= V/r

the Bulb B will have current IB = IA- IC , so a lower glow than A.

and C and D will be equally glowing but lower than A as well as B.

moreover , when one removes D now V= IB.r +IC.r and IB= IC

so IB= IA/2

and B as well as C will glow equally less than A. and C will be brighter than the previous arrangement of C and D parallel.

Though its true that in parallel bulbs arrangement removal of a bulb does not lead to any fluctuation in the other parallel connected bulb's brightness.

However, in your present arrangement the C and D are parallel not to voltage supply but is in series with B and sharing the input voltage with B. that's why this seemingly opposite fluctuation is being seen.

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### QuestionGuy

A guy learning physics.

Updated on November 07, 2022

In my book it reads, "The brightness of C: Increase. (If you remove D) Previously C was receiving only half of the current through the lower branch, now C is receiving all of the current coming out of B."

This makes sense, but everything that I found online says that C would stay the same brightness.

This is how I would try to solve it. The voltage and the resistance do not change, and using I = V/R, the current does not change and therefore the current does not change.

• The Photon over 5 years
Are you sure the situations where it was claimed to stay the same brightness also had a third bulb in position "B"?
• QuestionGuy over 5 years
The other scenarios only use C and D. Would B affect C and D?
• The Photon over 5 years
Of course it will. If you just have the voltage source and C and D, you know the exact voltage across C & D right away. If you include B, then the voltage drop across B affects the voltage across C & D (and depends on the resistance of C & D).
• QuestionGuy over 5 years
(A, B, C, and D are identical) I still am a little confused. The path that the current takes can be B --> D or B --> C. Isn't the voltage and resistance equal for both of them?
• The Photon over 5 years
Have you learned Kirchoff's Current Law? The current that flows through B has to be split between C and D. Remove D and then all the current through B must also flow through C.
• The Photon over 5 years
Or if you are only just learning resistor combinations, find the equivalent resistance of B, C, & D, vs the just B & C. You should find a different value, so a different total current flows from the battery through the bulbs depending on the configuration.
• QuestionGuy over 5 years
Yes, but if you remove B, doesn't it still have to split?
• The Photon over 5 years
You mean if you short B? Then it splits, but the voltage source adjusts its current to keep its output voltage fixed.
• The Photon over 5 years
If you know KCL and KVL, just assume some value for the bulb resistance and solve the three different circuits. If you don't know those rules yet, that's the next thing you should study.
• John Rennie over 5 years
Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems.
• robert bristow-johnson over 5 years
if the output impedance of the battery were zero, then removing A would not change the brightness of the other lights. but the internal battery resistance is not zero, so i imagine that B, C, and D will burn a little bit brighter with A removed.
• sammy gerbil over 5 years
This is how I would try to solve it. The voltage and the resistance do not change... If you remove a resistor, how can the resistance not change?