TrojanPoem
  • TrojanPoem
Find R if Pw on it was 20 watt
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TrojanPoem
  • TrojanPoem
|dw:1432754744584:dw|
TrojanPoem
  • TrojanPoem
@Michele_Laino
Michele_Laino
  • Michele_Laino
is 20 Watt the power dissipated on R?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Michele_Laino
  • Michele_Laino
do you know how compute the equivalent resistance of your circuit?
TrojanPoem
  • TrojanPoem
yeah
Michele_Laino
  • Michele_Laino
what is that resistance?
TrojanPoem
  • TrojanPoem
The total resistance it not given
Michele_Laino
  • Michele_Laino
we can compute the total resistance
TrojanPoem
  • TrojanPoem
Will be exhausting much.. think.
TrojanPoem
  • TrojanPoem
(40+R)* 30 / (70+R) + 5
Michele_Laino
  • Michele_Laino
here are the necessary steps: |dw:1432755438420:dw|
TrojanPoem
  • TrojanPoem
(40+R)* 30 / (70+R) + 5
Michele_Laino
  • Michele_Laino
that's right!
TrojanPoem
  • TrojanPoem
Okie, now ?
Michele_Laino
  • Michele_Laino
I got this: \[\Large {R_{TOTAL}} = \frac{{1550 + 35R}}{{70 + R}}\]
TrojanPoem
  • TrojanPoem
You got rid of 5 ? by making them on one fraction ?
TrojanPoem
  • TrojanPoem
Well, Now what to do ?
Michele_Laino
  • Michele_Laino
yes!
Michele_Laino
  • Michele_Laino
we can compute the current, like below: \[\Large I = \frac{V}{{{R_{TOTAL}}}} = \frac{{\left( {70 + R} \right)V}}{{1550 + 35R}}\]
TrojanPoem
  • TrojanPoem
Ok , Now ?
Michele_Laino
  • Michele_Laino
|dw:1432755794918:dw|
Michele_Laino
  • Michele_Laino
we can compute the voltage drop V_AB such that: |dw:1432755868980:dw| \[\Large {V_{AB}} = V - 5I = V - \frac{{5\left( {70 + R} \right)V}}{{1550 + 35R}}\]
Michele_Laino
  • Michele_Laino
please, simplify that expression
Michele_Laino
  • Michele_Laino
whatdo you get?
Michele_Laino
  • Michele_Laino
oops..what do you get?
TrojanPoem
  • TrojanPoem
- 26250 - 375R / 1550 + 35 R + V
TrojanPoem
  • TrojanPoem
HUGE NUMBER !
Michele_Laino
  • Michele_Laino
I got this: \[\Large \begin{gathered} {V_{AB}} = V - 5I = V - \frac{{5\left( {70 + R} \right)V}}{{1550 + 35R}} = \hfill \\ \hfill \\ = \frac{{\left( {1200 + 30R} \right)V}}{{1550 + 35R}} \hfill \\ \end{gathered} \]
TrojanPoem
  • TrojanPoem
Is V the VB or V at 5 ohm resistor ?
Michele_Laino
  • Michele_Laino
it is V_AB
TrojanPoem
  • TrojanPoem
VAB = the current X the total resistance between AB
TrojanPoem
  • TrojanPoem
Or VB - V at 5 ohm resistor Or VAB = VA - VB
Michele_Laino
  • Michele_Laino
|dw:1432756298125:dw|
TrojanPoem
  • TrojanPoem
Well, now what will we do ?
Michele_Laino
  • Michele_Laino
we can compute the current along the resistors 40 and R, namely the current I_2: |dw:1432756485462:dw|
TrojanPoem
  • TrojanPoem
.... ?
Michele_Laino
  • Michele_Laino
so we can write: \[\Large \begin{gathered} {I_2} = \frac{{{V_{AB}}}}{{40 + R}} = \frac{{\left( {1200 + 30R} \right)V}}{{1550 + 35R}} \times \frac{1}{{40 + R}} = \hfill \\ \hfill \\ = \frac{{30V\left( {40 + R} \right)}}{{\left( {1550 + 35R} \right)\left( {40 + R} \right)}} = \hfill \\ \hfill \\ = \frac{{30V}}{{1550 + 35R}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
what do you think about that?
TrojanPoem
  • TrojanPoem
PW = I^2 R solve for R ,
Michele_Laino
  • Michele_Laino
yes! Nevertheless we don't know the value of R
Michele_Laino
  • Michele_Laino
Next we can compute the value of the voltage drop on resistor R, as below:
Michele_Laino
  • Michele_Laino
|dw:1432756998788:dw|
TrojanPoem
  • TrojanPoem
(30V / 1550 + 35 R)^2 * R = 20 30V/ 1550 + 35R * sqrt(R) = 2sqrt(5)
TrojanPoem
  • TrojanPoem
30 * 75 sqrt(R) = 2 sqrt(5) (1550 +35R)
Michele_Laino
  • Michele_Laino
\[\large {V_1} = {V_{AB}} - 40{I_2} = \frac{{\left( {1200 + 30R} \right)V}}{{1550 + 35R}} - \frac{{40 \times 30V}}{{1550 + 35R}} = ...\]
Michele_Laino
  • Michele_Laino
|dw:1432757229841:dw|
TrojanPoem
  • TrojanPoem
2250 sqrt(R) = 31500sqrt(5) + 70sqrt(5)R 70sqrt(5)R - 2250sqrt(R) - 31500sqrt(5) = 0 R = 15 sqrt(437) + 45sqrt(5) / 14
TrojanPoem
  • TrojanPoem
You miscalculated the current :(( noooooooooooooooooooooooooooooooooooo
Michele_Laino
  • Michele_Laino
no, my computations are correct!
Michele_Laino
  • Michele_Laino
I got: \[\Large {V_1} = \frac{{30RV}}{{1550 + 35R}}\]
TrojanPoem
  • TrojanPoem
Now, how will we get R ?
Michele_Laino
  • Michele_Laino
we have to use the power dissipated by R, so we can write this formula: \[\Large W = \frac{{V_1^2}}{R}\]
Michele_Laino
  • Michele_Laino
where W=20 Watt
Michele_Laino
  • Michele_Laino
after a simple substitution, we get: \[\Large W = \frac{{V_1^2}}{R} = \frac{{900R{V^2}}}{{{{\left( {1550 + 35R} \right)}^2}}}\]
Michele_Laino
  • Michele_Laino
so we got this condition: \[\Large 20 = \frac{{900R{{\left( {75} \right)}^2}}}{{{{\left( {1550 + 35R} \right)}^2}}}\]
Michele_Laino
  • Michele_Laino
which is a quadratic equation for R. Please solve it for R
TrojanPoem
  • TrojanPoem
20 ( 1550 + 35R)^2 = 900R(75)^2 2402500 + 24500R^2 + 1085000R = 5062500R 24500R^2 -3977500 R + 2402500 = 0 R = 161, 7 R= 0.7
TrojanPoem
  • TrojanPoem
hmm , Is that normal ?
Michele_Laino
  • Michele_Laino
please, wait I'm checking your computation
TrojanPoem
  • TrojanPoem
Take your time Pr0
TrojanPoem
  • TrojanPoem
Hey
Michele_Laino
  • Michele_Laino
I got these values: \[\Large \begin{gathered} R = \frac{{5365 + 2475}}{{98}} = 80\;Ohms \hfill \\ \hfill \\ R = \frac{{5365 - 2475}}{{98}} \cong 29.5\;Ohms \hfill \\ \end{gathered} \]
TrojanPoem
  • TrojanPoem
Thanks so much.
Michele_Laino
  • Michele_Laino
thanks! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.