RE: Eletrical Circuit Question 05-07-2015, 05:31 AM
#31
Unless the copper you are putting in to replace bulb 7 is an ideal short (0 ohms, and impossible thanks to internal resistance) then all bulbs will still be receiving power. The lower the resistance of this short the less current the other parallel branches will see (it may not even be enough for the bulbs to light) but it WILL spread out according to Current Dividing.
For the record the current in a given branch is found by It(Rt/Rb+Rt)
Where It = Total Current, and Rb and Rt are the Resistance of the branch and total resistance of all the branches respectively.
So even if the other bulbs dont light up there will still be power across them.
Feel free to check this using a multimeter
For the record the current in a given branch is found by It(Rt/Rb+Rt)
Where It = Total Current, and Rb and Rt are the Resistance of the branch and total resistance of all the branches respectively.
So even if the other bulbs dont light up there will still be power across them.
Feel free to check this using a multimeter
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