Hi im trying to find the total impedance in this circuit but im having troubles figuring it out. I have calculated the individual impedance for each branch like following:
Xc = 46.8 ohm at -90 degrees
Xl = 46.8 ohm at 90 degrees
ZR = 1000 + j0
Zc = 0 – j46.8
ZRsL = 10 + j46.8 => (47.8 ∠ 77.93)
So now when i try to find the total impedance the first thing i tried was ZT = 1/((1/ZR)+(1/Zc)+(1/ZRsL)) which i get to be 23.1 ohm. Which is wrong. Any help would be appreciated.
EDIT: So i get how do do it from the comments below but i dont know what numbers to use. Like with (ZT=(Z1⋅Z2⋅Z3)/(Z1⋅Z2+Z1⋅Z3+Z2⋅Z3)) What do i actually put as Z1, Z2 etc. I cant take (10 + j46.8)( 0 – j46.8) can i? In my mind that is 56.8-46.8. I know it isnt but i cant figure out how to do it
Best Answer
I think an earlier version of your question suggested that you may have an answer, just not a process for finding it. If so, you might have posted it up.
Ignoring the voltage source, except for the frequency, I get about the same figures you do for each branch:
Putting those in parallel, leads to something like:
\$1000\Omega \vert\vert (10\Omega + 14.9mH ) \vert \vert 6.8\mu F = 180.943269 - j31.4410821\$
or,
\$183.654589 \angle -9.85741212^{\circ}\$
I'm just not sure how you got your figure. Do you know if I got my calculations right, though, according to what you were told? I'm reasonably confident about my figure here. But I make mistakes, too.
EDIT: I just used the same thing you referred to:
\$Z_T = \frac{1}{\frac{1}{Z_1} + \frac{1}{Z_2} + \frac{1}{Z_3}} = \frac{Z_1 \cdot Z_2 \cdot Z_3}{Z_1 \cdot Z_2 + Z_1 \cdot Z_3 + Z_2 \cdot Z_3}\$
I used the complex number notation, because it's easier for me. But the above is just the same as:
\$Z_T = \frac{1}{\frac{1}{1000} + \frac{1}{10 + j46.8097305} + \frac{1}{-j46.8102774}}\$
EDIT AGAIN: Note to @Joo223: If you want to test out either form of the equation I mentioned, then you MUST keep the fully cartesian equivalent. This means keeping your \$\sqrt{-1}\$ terms separated out. Remember, these are complex numbers and you are using a complex number system. Not the real number system. When you compute Z for something, you get a complex value. The imaginary part might be 0. The real part might be zero. But you still can't just go around thinking that an imaginary part is really just a real part. It's not. You will need a complex number calculator.
So look at these expressions:
\$\frac{1}{\frac{1}{1000}+\frac{1}{10 + 46.8097305 i}+\frac{1}{-46.8102774 i}}\$
or, equivalently,
\$\frac{1000 \cdot (10 + 46.8097305 i) \cdot (-46.8102774 i)}{1000 \cdot (10 + 46.8097305 i) + 1000 \cdot (-46.8102774 i) + (10 + 46.8097305 i) \cdot (-46.8102774 i)}\$
Please take careful note that I've not just ignorantly stuffed a real numbered value as an \$\Omega\$ value. I've had to keep track of both real and imaginary terms. You will need to use complex number calculations here.
Real numbers do not form a closed algebra over their operations. Complex numbers do close the algebra. But more importantly for electronics is the idea that multiplication by i rotates a plot by \$90^{\circ}\$. Or, you might look up Euler's: \$ e^{i \cdot \theta} = cos(\theta) + i\cdot sin(\theta)\$. Investigate the infinite series for e, sine, cosine, and the hyperbolic sine and cosine. It may help, too.
But the point is that you must keep things straight in your calculations. I have a complex number calculator that handles everything smoothly for me. So I use it. I can set up the \$\omega = 2\cdot\pi\cdot f\$, as well, and just slap that in along the way. In fact, my calculator allows me to simply type this into it: \$1000 \vert (10 + 14.9mH ) \vert 6.8\mu F\$ and to get the right answer without any further ado. It knows about H and it knows about F and it knows about SI. It knows that + means in series and that | means in parallel. So it's really easy for me to make sure I'm getting the right results. (I created that calculator software for exactly these reasons.)