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I was trying to resolve the following problem:

A 6.8 kΩ resistor, a 7 mH coil, and a 0.02 uF capacitor are in parallel across a 17 kHz ac source. The coil's internal resistance, RW, is 30 Ω.

The equivalent parallel resistance, Rp(eq) is

But, the correct of the exercise says it's 18 Ω, and this answer doesn't make sense to me.

Actually, to find the answer, I calculated the sum of admittances Y then I expressed R (the real part of the impedance Z) as R = G/G^2+B^2 with G being the conductance et B the susceptance. I started by calculating XL and XC then I calculated G and B and finally applied this formula R = G/G^2+B^2 that I got starting from the fact that we know that Z = 1/Y

By doing that, I found that the equivalent resistance is equal to 294.2 Ω which is far from 18 Ω.

Would you have any clues of why the correction says that the answer is 18 Ω? This really makes no sense to me.

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  • \$\begingroup\$ The value of C should be ~ 0.5uf ... \$\endgroup\$ Commented Feb 15 at 13:38
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    \$\begingroup\$ @Antonio51 If you're assuming resonance, you've forgotten the 2pi, but the question doesn't claim resonance anyway \$\endgroup\$ Commented Feb 15 at 14:38
  • \$\begingroup\$ No. Not resonance ... but after, at 17 kHz. \$\endgroup\$ Commented Feb 15 at 15:22

3 Answers 3

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It's likely an error in the question or answer to it. Even the capacitor alone has only 470 ohms impedance at 17 kHz, there is no way the answer drops to 18 ohms.

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Well, the resistance can be found using:

$$\vec{\text{Z}}_\text{i}=\left(\frac{\displaystyle1}{\displaystyle\text{R}}+\frac{\displaystyle1}{\displaystyle\text{R}_\text{w}+\text{j}\omega\text{L}}+\frac{\displaystyle1}{\displaystyle\frac{1}{\text{j}\omega\text{C}}}\right)^{-1}=\left(\frac{\displaystyle1}{\displaystyle\text{R}}+\frac{\displaystyle1}{\displaystyle\text{R}_\text{w}+\text{j}\omega\text{L}}+\text{j}\omega\text{C}\right)^{-1}\tag1$$

Using your values, we get:

$$\left|\vec{\text{Z}}_\text{i}\right|\approx1211.03\space\Omega\tag2$$

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  • \$\begingroup\$ The question is asking for the equivalent resistance, not the impedance. I get \$Z_i = 294.3 - 1175j\$ which is consistent with the OP's resistance value. \$\endgroup\$ Commented Feb 15 at 19:08
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You are asked to find RP, not RS.
You started out in the right direction of using admittance to solve the problem but switched over to impedance which leads you to finding RS, not RP.

At 17 kHz:
Y = 2.00635e-4 + j8.0099e-4

Thus, RP = 1/real(Y) = 4984.2 ohms

Where the answer guide got 18 ohms is a mystery. It's not the first time answer guides are wrong.

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