The solutions are, 2.677A and 0.523A.
For the case of 2.677A, the resistors are 5Ohms + 1.120Ohms + 0.977Ohms
For the case of 0.523A the resistors are 5Ohms + 5.737Ohms + 25.595Ohms
Two different circuits satisfied by the same equation.
I thought I already answered this, but I can't find it. Did you delete it? Not good.
Again, you do not have two currents at the same time, or with the same circuit for that matter.
You have two different circuits, with two different currents, which will satisfy the conditions of the problem. Note that the two different circuits have different resistor values.
edit, interestingly, I found the other answer, and it is identical with this one but under a different account name. Below is my answer. Note that using two accounts this way is not allowed.
It's 5 ohm, or 5 Ω, NOT ohm's
let x be total current
x = Rt/19
for the third resistor, P = EI
E = 7/x
19 = 5x + 3 + 7/x
5x + 7/x – 16 = 0
5x2 – 16x + 7 = 0
solutions are x = 2.68, 0.523
let's try I = 2.68 amp
voltage across 5 Ω is 13.4 volts
3 volts across the middle R
and for the third, it has 19–13.4–3 = 2.6 volts, and with 2.68 amp, that is 6.97 watts, close enough
let's try I = 0.523 amp
voltage across 5 Ω is 2.615 volts
3 volts across the middle R
and for the third, it has 19–2.615–3 = 13.385 volts, and with 0.523 amp, that is 7.00 watts, close enough
so there are indeed two solutions
you don't have two currents, you have two possible answers for a complex problem.
Yes it is possible to have two different circuits with different values of current and yet the same value source Voltage. But R2 and R3 must have a different resistance value in each circuit to satisfy the given 3V across R2 , the given 7 Watts dissipated in R3 and the given 5 Ohm value for R1.
Set the circuit Watts equal to the source Watts.
(i^2*R1) + i*(V of R2) + (i^2*R3) = I*(V of source)
Plug in the all the given values and get;
5i^2 + 3i + 7 = 19i
5i^2 -16i + 7 = 0
Solve for i and get;
i = 2.697 Amps or .523 Amps
In the circuit where R2 and R3 have resistance values that make i total = 2.697 Amps then;
R2 must = 3V/2.697A = 1.112 Ohm
R3 must = 7W/(2.697A^2) = .962 Ohm
But in the circuit where R2 and R3 have resistance values that make i total = .523 Amps then;
R2 must = 3V/.523A = 5.736 Ohm
R3 must = 7W/(.523A^2) = 25.591 Ohm
You don't need a quadratic for this. It is the same everywhere, and is given by 19/(5+3+7)=1.267 amps.
19 Voltages source, R1 = 5 ohm's, 3 volt is the voltages across R2 and 7 watt's across R3, connected in series circuits? i solve it using quadratic equation that given me a two answer CURRENT 1 and CURRENT 2, that's why i ask if it can be possible to have to current in series with one source voltage??
http://www.facebook.com/photo.php?fbid=451950588189954&set=a.451950454856634.117934.100001250505771&type=1&theater
this the diagram of that circuit
