I think I am not doing the algebra correctly when using the-b-+root of b2-4ac/2a
ID: 1662631 • Letter: I
Question
I think I am not doing the algebra correctly when using the-b-+root of b2-4ac/2a. Thank.Explanation / Answer
Let the current in the circuit on the left be I1 Let the current in the circuit on the right be I2 Circuit on the left: -86 + 5I1 + 30(I1 - I2) = 0 => 35I1 - 30I2 = 86 .............(1) Circuit on the right: 30(I2 - I1) + 40I2 + RI2 = 0 => -30I1 + (70 + R)I2 = .............. (2) We know that P = I2R => 16 = I22R => R = 16/I22 .............(3) Substitute (3) into (2) -30I1 + (70 + 16/I22)I2 = 0 =>-30I1 + 70I2 + 16/I2 = ............. (4) Now, solve (1) and (4) to get the value of I2 Once you have I2, plug it back into (3) and solve for R I guess the best way to solve this equation is to solve (1) forI1 35I1 - 30I2 = 86 35I1 = 30I2 + 86 I1 = (86 + 30I2)/35 Plug this into (2) -30I1 + 70I2 + 16/I2 = 0 -30[(86 + 30I2)/35] + 70I2 + 16/I2 = 0 Common denominator is 35I2 (-30)(86 + 30I2)(I2) + 35I2(70I2) + (35)(16) = 0 -2580I2 - 900I22 + 2450I22 + 350 = 0 1550 I22 - 2580I2 + 350 = 0 I2 = ....... A or I2 = ......... A Plug this into (3) R = 16 / (I2)2 = .........O or R = 16 / (I2)2 = ........O