Find R if Pw on it was 20 watt

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Find R if Pw on it was 20 watt

Physics
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|dw:1432754744584:dw|
is 20 Watt the power dissipated on R?

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Other answers:

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

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