# Simple Circuit Solving Help!

#### Musfikur Rahman

Joined Feb 17, 2016
5
Can anyone solve this problem to find the current in the circuit please? I need it badly.

#### WBahn

Joined Mar 31, 2012
30,062
So do you just want someone to solve it and give you the answer?

#### Musfikur Rahman

Joined Feb 17, 2016
5
So do you just want someone to solve it and give you the answer?
No, I just don't need the 'answer', I need a sollution step by step. I tried to solve this in many ways, but it always ends with a quardatic equation whitch is 15i^2-23i+16=0. I am pissed off & I posted it here.

#### KL7AJ

Joined Nov 4, 2008
2,229
Can anyone solve this problem to find the current in the circuit please? I need it badly.

#### WBahn

Joined Mar 31, 2012
30,062
Is there ANY chance we might talk you into actually showing YOUR work so that we can help you see where YOU went wrong?

Or do you want someone to do ALL of your work for you?

#### KL7AJ

Joined Nov 4, 2008
2,229
Can anyone solve this problem to find the current in the circuit please? I need it badly.

#### Musfikur Rahman

Joined Feb 17, 2016
5
So if a Resistance value and a Power value given in series with a input voltage, it is not possible to find the circuit current?

#### WBahn

Joined Mar 31, 2012
30,062
No, I just don't need the 'answer', I need a sollution step by step. I tried to solve this in many ways, but it always ends with a quardatic equation whitch is 15i^2-23i+16=0. I am pissed off & I posted it here.
What's wrong with getting a quadratic equation?

#### WBahn

Joined Mar 31, 2012
30,062
That problem is not solvable with the information given.
Yes and no. The given information is sufficient, it's just that no solution exists for those particular values. The most total power than can be delivered to the set of parallel resistors is a tad under 90 W.

#### Musfikur Rahman

Joined Feb 17, 2016
5
What's wrong with getting a quadratic equation?
The answer will come with complex form. Please just confirm me my steps are right, or wrong.

#### WBahn

Joined Mar 31, 2012
30,062
So if a Resistance value and a Power value given in series with a input voltage, it is not possible to find the circuit current?
That's not the problem. If the powers given had each been, say, 1/2 of what they presently are, the solution would be easy to find.

Look at it this way -- let's say that instead of two resistors with the power being dissipated you had been given a single resistor with the claim that the power being dissipated is one million watts. Is that possible, given the presence of the 150 Ω resistor? If not, what is the absolute maximum amount of power that a 230 V source having a 150 Ω series resistance can possible dump into another resistance.

#### WBahn

Joined Mar 31, 2012
30,062
The answer will come with complex form. Please just confirm me my steps are right, or wrong.
Even if we ignore units, that last line is not consistent with the line above it.

(15 Ω)·I² - (23 V)·I + (16 W) = 0

If we divide by sides by 1 Ω, we have

(15)·I² - (23 A)·I + (16 A²) = 0

#### KL7AJ

Joined Nov 4, 2008
2,229
Yes and no. The given information is sufficient, it's just that no solution exists for those particular values. The most total power than can be delivered to the set of parallel resistors is a tad under 90 W.
Oh yeah...you're right. It is a doozy of a problem though!

#### WBahn

Joined Mar 31, 2012
30,062
If you get a complex number for the current through a resistor, what does that tell you about the problem?

#### WBahn

Joined Mar 31, 2012
30,062
The answer will come with complex form. Please just confirm me my steps are right, or wrong.
Oh, and thank you for showing your work. Now it is obvious that you are NOT just looking for someone to do your work for you and that you HAVE actually put in some quality effort. We are now in a much better place to help you make sense of the problem and why no solution exists, as stated.

#### WBahn

Joined Mar 31, 2012
30,062
Oh yeah...you're right. It is a doozy of a problem though!
Depending on how you approach it, it can be either a brain twister or it can be easy. I started down the brain twist path initially, but backed off and came at it another way. I should have looked at the maximum power transfer possible at that point, but didn't until after my second way, which involves solving a quadratic that can be trivially written down by inspection, yielded no real solutions and I wanted to understand why not. The TS took a slightly different route than I did, but still used an approach that is very commendable.

#### KL7AJ

Joined Nov 4, 2008
2,229
Depending on how you approach it, it can be either a brain twister or it can be easy. I started down the brain twist path initially, but backed off and came at it another way. I should have looked at the maximum power transfer possible at that point, but didn't until after my second way, which involves solving a quadratic that can be trivially written down by inspection, yielded no real solutions and I wanted to understand why not. The TS took a slightly different route than I did, but still used an approach that is very commendable.
You can always add a negative resistance.

#### WBahn

Joined Mar 31, 2012
30,062
You can always add a negative resistance.
Works for me! Does Radio Shack still carry them thar negisters?

#### Musfikur Rahman

Joined Feb 17, 2016
5
Even if we ignore units, that last line is not consistent with the line above it.

(15 Ω)·I² - (23 V)·I + (16 W) = 0

If we divide by sides by 1 Ω, we have

(15)·I² - (23 A)·I + (16 A²) = 0
Track the units is the right way to write down an equation? This is new to me & it is not generally practiced here. By the way thanks for your effort to this problem

#### shteii01

Joined Feb 19, 2010
4,644
I got a question about those parallel resistors. When OP simplified them, he simply added the power of one resistor to the power of the other resistor. Is that correct way to do it for the parallel resistors?