Calculating Voltage Difference in a Parallel and Series Circuit

AI Thread Summary
The discussion focuses on calculating the voltage difference across the third bulb in a circuit with three 60W, 120V bulbs, where two are in parallel and one is in series with a 120V battery. The total circuit resistance is determined by combining the resistance of the parallel and series configurations. The resistance of each bulb is calculated using R = V^2/P, leading to a total resistance of 360 ohms. The current through the circuit is found to be 0.33A, resulting in a voltage drop of 80V across the third bulb, which is noted as negative due to the loss of voltage. The calculations and reasoning provided guide towards completing the problem independently.
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Homework Statement


Three 60 w 120 V bulbs are connected to a 120 V battery. 2 Bulbs are in parallel with each other in a smaller loop. The third is in series with the battery and the other loop.
find voltage difference across bulb three.


Homework Equations


R= V^2/P
1/R_loop= 1/R_1


The Attempt at a Solution


R=120^/60

1/R_total = 2/240 R_total= 120 +240 =360

I _total = V/R = 120/360
 
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Start by finding the total circiut resistance.

You must have RT= RB+ RL

Were:
RB is the resistance of a single bulb and
RL is the resistance of the parallel pair of bulbs.

RL is :

\frac 1 {R_L} = \frac 1 {R_B} + \frac 1 {R_B}

So RL = .5 Rb

Now you should be able to complete the problem on your own.
 
I got V_3 is .33A * 240 = 80v but its -80v b/c its loosing right?
 

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