Asked by Azizur
Two bulbs are connected in parallel across a source of emf E = 11.0 V with negligible internal resistance. One bulb has a resistance of 3.0 Ω, and the other is 2.5 Ω. A resistor R is connected in the circuit in series with the two bulbs. What value of R should be chosen in order to supply each bulb with a voltage of 2.4 V? For what value of R would the potential difference across each of the bulbs be 2.4 V?
Answers
Answered by
Physics needed ---- Reiny
I am not up on my physics, last time I studied physics was 63 years ago.
Answered by
oobleck
The two bulbs in parallel produce an equivalent resistance X such that
1/X = 1/3 + 1/2.5
So, X = 1.36 Ω
Now you just have an equivalent series circuit with two resistors, R and 1.36Ω. So now you can probably figure out how to find R so that you get a 2.4V drop across 1.36Ω, yeah?
1/X = 1/3 + 1/2.5
So, X = 1.36 Ω
Now you just have an equivalent series circuit with two resistors, R and 1.36Ω. So now you can probably figure out how to find R so that you get a 2.4V drop across 1.36Ω, yeah?
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