Calculating Voltage Difference in a Parallel and Series Circuit

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SUMMARY

The discussion focuses on calculating the voltage difference across a bulb in a circuit configuration involving three 60W, 120V bulbs connected to a 120V battery. Two bulbs are arranged in parallel, while the third is in series with the battery. The total circuit resistance is determined to be 360 ohms, leading to a calculated current of 0.33A. The voltage across the third bulb is found to be 80V, with a negative sign indicating a voltage drop.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Knowledge of series and parallel circuit configurations
  • Familiarity with calculating resistance in parallel circuits
  • Ability to manipulate electrical power equations (P = V^2/R)
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  • Learn to calculate total resistance in complex circuits
  • Explore the implications of voltage drops in series circuits
  • Investigate the effects of power ratings on circuit design
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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 :

[tex]\frac 1 {R_L} = \frac 1 {R_B} + \frac 1 {R_B}[/tex]

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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