Desperately need introductory CIRCUIT/RESISTOR help

  • Thread starter Thread starter martinlematre
  • Start date Start date
  • Tags Tags
    Introductory
Click For Summary
SUMMARY

This discussion revolves around creating a circuit with three 10.0Ω light bulbs powered by a 12.0V battery, where the brightness of the bulbs must be in the ratio of 1:2:3. The participants clarify that brightness is defined as power, calculated using the formula P = V²/R. The challenge lies in configuring 10.0Ω resistors to achieve the desired brightness ratios, particularly in parallel circuits where voltage remains constant. The conclusion emphasizes the need for creative resistor configurations to approximate the required power outputs.

PREREQUISITES
  • Understanding of Ohm's Law and power calculations (P = V²/R)
  • Knowledge of series and parallel circuit configurations
  • Familiarity with resistor values and their impact on circuit behavior
  • Basic circuit design skills, particularly with light bulbs and resistors
NEXT STEPS
  • Explore advanced circuit design techniques for achieving specific power ratios
  • Learn about the behavior of resistors in parallel and series configurations
  • Investigate the relationship between current (I) and brightness in circuits
  • Utilize circuit simulation software to visualize and test different configurations
USEFUL FOR

Students studying electrical engineering, hobbyists building circuits, and educators teaching concepts of electricity and power in circuits.

martinlematre
Messages
41
Reaction score
0

Homework Statement


You’re given a 12.0V battery, 3 light bulbs (of 10.0Ω each), and LOTS of 10.0Ω resistors. Make a circuit so that the light bulbs’ brightnesses are multiples of 1, 2, 3. In other words, if the dimmest light bulb has brightness B, the second dimmest will have brightness 2B, and the brightest will be 3B.

Homework Equations



I've derived a few over a few hours and have drawn it out and mapped and graphed v^2/r (I'm assuming in this case brightness means POWER which is vsquared over r)

The Attempt at a Solution


I just don't see how this is possible to actually map out these 10 ohm resistors with a 10 ohm bulb and a set 12 volt battery. I am under the presumption that by brightness my teacher means power, and having graphed the voltages after crossing a list of quantities of 10 ohm resistors, I cannot find 3 powers that are direct multiples of 2 and 3 of one resistor configuration.

Note that there is no statement for what we are defining as brightness, we have not been taught anything about brightness and that's actually the first time the word has popped up in this course. I don't understand because under my understanding, anything you can construct on a resistor circuit is not linearly related to the brightness, but rather the measurement of lumens of the bulb. Would appreciate if anoyne helped
 
Last edited:
Physics news on Phys.org
hi martinlematre! :smile:

hint: parallel :wink:
 
tiny-tim said:
hi martinlematre! :smile:

hint: parallel :wink:

I am using parallel circuits. The powers (V^2 over Resistors) are not linearly scaling with 10 ohm resistors
 
?? :confused:

V will be the same for parallel circuits
 
tiny-tim said:
?? :confused:

V will be the same for parallel circuits

There's a voltage drop across a resistor whose specific quantity is proportional to power (What I'm assuming is what my teacher means by brightness).

V^2/R will be the variable Voltage after drops (And before each lightbulb) squared, over 10ohms (The lightbulb.

Except there is no configuration where that variable voltage after going through X resistors has a 1, 2x and 3x value for power
 
martinlematre said:
Except there is no configuration where that variable voltage after going through X resistors has a 1, 2x and 3x value for power

the question says "LOTS" of resistors, so i think you're expected to get only approximately 1:2:3 :wink:

(and it would be more logical to use I, not V)
 
tiny-tim said:
the question says "LOTS" of resistors, so i think you're expected to get only approximately 1:2:3 :wink:

(and it would be more logical to use I, not V)

except I am expected to draw/use a very limited program where i can't fit more than 5 or 6 resistors in a circuit without it crashing.

im just making brightness proportional to I.
 
Suppose you had 3 light bulbs in series across the 12 volt battery.

Then you rewire the circuit to have 1 light bulb in series with two 10 ohm resistors.

Would the first circuit be exactly 3 times as bright as the second?
 

Similar threads

Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
3
Views
2K
Replies
7
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 45 ·
2
Replies
45
Views
4K
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
15K