# Homework Help: Help with circuit

1. Jun 1, 2013

### jafferrox

1. The problem statement, all variables and given/known data

2. Relevant equations
XL = wL
XC = 1/wC

3. The attempt at a solution
XL = 4pi

2. Jun 1, 2013

### Staff: Mentor

The first step is to determine the resultant impedance of the two parallel branches. There should be an imaginary operator, j, included in your impedance term for the L and C elements.

3. Jun 1, 2013

### jafferrox

and then what's the next step?

4. Jun 2, 2013

### Staff: Mentor

Step 2 is to equate the impedance to be a pure resistance, because at resonance a circuit appears purely resistive to any applied voltage.

5. Jun 2, 2013

### jafferrox

doesn't this mean XL=XC, i already found XL and XC is the same and then solve for c?

6. Jun 2, 2013

### Staff: Mentor

It does, where XL and Xc are the only impedances present. But there is an R here, and that makes things more interesting.

That gives a rough approximation, and will usually get you near the right answer, but can be wrong by up to about 15%.

7. Jun 2, 2013

### anhnha

I think you need to find impedance in the form: Z= f(R) + j*f(R, XL, XC).
Then solve f(R, XL, XC) = 0 for C with R, XL and W given.

8. Jun 3, 2013

### jafferrox

I don't know what you mean, can you please make it clearer?

Thanks

9. Jun 3, 2013

### anhnha

Can you get the impedance of two parallel branches?
Using ZL = jXL and ZC = -jXC
The first branch: Z1= ZC = -jXC
The second branch: Z2 = R + jXL
Then total impedance:
1/Z = 1/Z1 + 1/Z2
You need to solve for Z and then force imaginary part of Z equals zero. You will find C.

10. Jun 3, 2013

### jafferrox

I couldn't do it, can someone please do it and attach a picture or a screenshot of the working out.

Thanks