Confusing resistance and power question

Watch
xx360noscoperxx
Badges: 3
Rep:
?
#1
Report Thread starter 4 months ago
#1
A circuit is set up with a cell of internal resistance r and e.m.f. V in series with a resistor of resistance R. The total power dissipated in the circuit is P1.

An identical cell is placed both in parallel with the first cell and in parallel with the resistor of resistance R, which are already in parallel with each other. The total power dissipated in the circuit is now P2. The ratio P2/P1 = 7/5.

What is the ratio R/r?

I got the equations: I = V2/r 2I = V1/R
P1 = V^2/(R+r) P2 = (2I)^2R + 2Ir
V = V1 + V2

I think my algebraic skills may be a bit off, or my equation for P2 must be wrong? The power dissipated from each circuit is the power dissipated from each resistor totalled up. I can't seem to simplify it down.

Thank you!
0
reply
Callicious
Badges: 22
Rep:
?
#2
Report 4 months ago
#2
Diagrams
Name:  1.PNG
Views: 13
Size:  19.1 KB
Name:  2.PNG
Views: 13
Size:  24.8 KB

Text
For the top circuit, the power is gotten via considering the power from each component, via
 \displaystyle P_1 = P_r + P_R = \frac{\left(\frac{r}{r + R}V\right)^2}{r} + \frac{\left(\frac{R}{r + R}V\right)^2}{R} = \frac{V^2}{r+R}
which is one of the equations you've got.

For the bottom circuit/second one, I'm having trouble visualizing what they're after, but the diagram I've drawn is the one I'll go on. The cells are in parallel, and then the resistor is slapped in "parallel" with them... although here it just seems to be in series with the two cells forming a parallel effective cell... anyway someone else reading this should definitely take a look and maybe try to help a bit better.

Anyway, going on the circuit I've drawn, you can see that the two cells are effectively in series with the resistor, forming an "effective cell" with a resistance

\frac{1}{r_{eff}} = \frac{1}{r} + \frac{1}{r}

which gives

r_{eff} = \frac{r}{2}

The new circuit effectively reduces to

Name:  3.PNG
Views: 11
Size:  22.1 KB

for that combination, which is something you can solve through in the exact same way as you did for the first circuit, except with \frac{r}{2} instead of r.
0
reply
Callicious
Badges: 22
Rep:
?
#3
Report 4 months ago
#3
To follow up a bit... the power in the new circuit is going to be something like
P_2 = \frac{V^2}{\frac{r}{2} + R}
and you know that
\frac{P_2}{P_1} = \frac{7}{5} = \frac{r+R}{\frac{r}{2}+R}
which is a perfectly solvable equation, at least to get r(R) or whatever.

Again please for the love of god check over what I've written, my interpretation of the question is likely off.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

What support do you need with your UCAS application?

I need help researching unis (27)
13.17%
I need help researching courses (14)
6.83%
I need help with filling out the application form (9)
4.39%
I need help with my personal statement (85)
41.46%
I need help with understanding how to make my application stand out (51)
24.88%
I need help with something else (let us know in the thread!) (3)
1.46%
I'm feeling confident about my application and don't need any help at the moment (16)
7.8%

Watched Threads

View All