TURN THE LIGHTS OFF..question

[discowisco]

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Three identical lightbulbs are connected to a battery and the lightbulbs are parallel. If the middle bulb burns out, what happens?

So I understand that the brightness remains the same since each bulbs Current is determined by I=V/R. However why does the total current through the circuit DECREASE?

The way I see it is that lets say the voltage provided by the battery is 100V and that each bulb provides a resistance of 5 ohms. Thus the current going through the circuit would be 100/ (5^3/15). Which would be 12. Now if there were only 2 bulbs working that would mean 100/(5^2/10) = 40. Thus with 2 bulbs isnt there more current in the circuit?

 

[LianaStan]

Adding resistors in a parallel manner will increase the total current that the battery can supply.

The Total Resistance = (1/R + 1/R + 1/R)^-1 for your specific example with 3 identical resistors. Keep in mind that you are doing inverses because it is a parallel circuit. Because of this, adding additional resistors in parallel will DECREASE total resistance and thus INCREASE total current as V=IR where V is constant.

As per you example, when all lightbulbs are working, Total Resistance = (1/R + 1/R + 1/R)^-1 = R/3. Thus, using V=IR, current will be 3I if we arbitrarily want the voltage to be equal to V.

Now when one lightbulb burns out, the total resistance is equal to (1/R + 1/R)^-1 which is R/2. Plugging this into V=IR, you get a current of 2I. As you can see, the total resistance increased and the total current decreased.

 

[discowisco]

Ic, but i thought adding resistors in parallel follow the formula - R1 X R2/ R1 + R2

 

[discowisco]

im guessing thats only for 2 resistors

 

[aspiringdoc09]

im guessing thats only for 2 resistors

Yes, that formula can only be used for 2 resistors (in parallel) or capacitors (in series) at a time. Honestly, it seems like the current should go up if you loose a resistor, but listen to the other person because physics always stomps me. But I know for sure my first statement is 100% accurate. lol

 

[LoLCareerGoals]

Easier way to think about is Kirkoff's law. Current into the node equals sum of currents out of the node.
Was: Inet = I1+I2+I3 = V/R1 + V/R2 +V/R3
Now: Inet_new = I1 + I2 = V/R1 + V/R2

Clearly Inet_new < Inet.