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TrojanPoem
 one year ago
Find R if Pw on it was 20 watt
TrojanPoem
 one year ago
Find R if Pw on it was 20 watt

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TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0dw:1432754744584:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1is 20 Watt the power dissipated on R?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1do you know how compute the equivalent resistance of your circuit?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1what is that resistance?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0The total resistance it not given

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can compute the total resistance

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Will be exhausting much.. think.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0(40+R)* 30 / (70+R) + 5

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here are the necessary steps: dw:1432755438420:dw

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0(40+R)* 30 / (70+R) + 5

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I got this: \[\Large {R_{TOTAL}} = \frac{{1550 + 35R}}{{70 + R}}\]

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0You got rid of 5 ? by making them on one fraction ?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Well, Now what to do ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can compute the current, like below: \[\Large I = \frac{V}{{{R_{TOTAL}}}} = \frac{{\left( {70 + R} \right)V}}{{1550 + 35R}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1432755794918:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we 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
 one year ago
Best ResponseYou've already chosen the best response.1please, simplify that expression

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1oops..what do you get?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0 26250  375R / 1550 + 35 R + V

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.0Is V the VB or V at 5 ohm resistor ?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0VAB = the current X the total resistance between AB

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Or VB  V at 5 ohm resistor Or VAB = VA  VB

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1432756298125:dw

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Well, now what will we do ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can compute the current along the resistors 40 and R, namely the current I_2: dw:1432756485462:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.1what do you think about that?

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0PW = I^2 R solve for R ,

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! Nevertheless we don't know the value of R

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1Next we can compute the value of the voltage drop on resistor R, as below:

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1432756998788:dw

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0(30V / 1550 + 35 R)^2 * R = 20 30V/ 1550 + 35R * sqrt(R) = 2sqrt(5)

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.030 * 75 sqrt(R) = 2 sqrt(5) (1550 +35R)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[\large {V_1} = {V_{AB}}  40{I_2} = \frac{{\left( {1200 + 30R} \right)V}}{{1550 + 35R}}  \frac{{40 \times 30V}}{{1550 + 35R}} = ...\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1432757229841:dw

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.02250 sqrt(R) = 31500sqrt(5) + 70sqrt(5)R 70sqrt(5)R  2250sqrt(R)  31500sqrt(5) = 0 R = 15 sqrt(437) + 45sqrt(5) / 14

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0You miscalculated the current :(( noooooooooooooooooooooooooooooooooooo

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, my computations are correct!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I got: \[\Large {V_1} = \frac{{30RV}}{{1550 + 35R}}\]

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Now, how will we get R ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have to use the power dissipated by R, so we can write this formula: \[\Large W = \frac{{V_1^2}}{R}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1after a simple substitution, we get: \[\Large W = \frac{{V_1^2}}{R} = \frac{{900R{V^2}}}{{{{\left( {1550 + 35R} \right)}^2}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so we got this condition: \[\Large 20 = \frac{{900R{{\left( {75} \right)}^2}}}{{{{\left( {1550 + 35R} \right)}^2}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1which is a quadratic equation for R. Please solve it for R

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.020 ( 1550 + 35R)^2 = 900R(75)^2 2402500 + 24500R^2 + 1085000R = 5062500R 24500R^2 3977500 R + 2402500 = 0 R = 161, 7 R= 0.7

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0hmm , Is that normal ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please, wait I'm checking your computation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I got these values: \[\Large \begin{gathered} R = \frac{{5365 + 2475}}{{98}} = 80\;Ohms \hfill \\ \hfill \\ R = \frac{{5365  2475}}{{98}} \cong 29.5\;Ohms \hfill \\ \end{gathered} \]
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