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This is a simple question on impedance and i just need to make sure i know the correct answer

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  • \$\begingroup\$ What did you calculate? \$\endgroup\$ – Andy aka Apr 6 '20 at 10:25
  • \$\begingroup\$ i calculated '3' ? \$\endgroup\$ – Mark Apr 6 '20 at 10:27
  • \$\begingroup\$ That's the wrong answer so what formula did you use. Feel free to find the formula using google. \$\endgroup\$ – Andy aka Apr 6 '20 at 10:29
  • \$\begingroup\$ Google the formula. Then plug the numbers in. This is help. \$\endgroup\$ – Andy aka Apr 6 '20 at 10:31
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    \$\begingroup\$ i used 1/ 2 (by)fc and got 15.915 kohm \$\endgroup\$ – Mark Apr 6 '20 at 10:39
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Well, we know:

$$\underline{\text{Z}}_\text{C}=\frac{1}{\text{j}\omega\text{C}}\tag1$$

And the reactance is given by:

$$\text{X}_\text{C}=\left|\frac{1}{\text{j}\omega\text{C}}\right|=\frac{1}{\left|\text{j}\omega\text{C}\right|}=\frac{1}{\omega\text{C}}\tag2$$

So, we get:

$$\text{X}_\text{C}=\frac{1}{2\pi\cdot1000\cdot0.01\cdot10^{-6}}=\frac{50000}{\pi}\approx15915.5\space\Omega\tag3$$

Which gives:

$$\text{X}_\text{C}\approx15.9155\space\text{k}\Omega\tag4$$

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    \$\begingroup\$ Whilst your answer is technically correct (arguably the best sort of correct) and beautifully formatted I suspect it's being downvoted because this is a homework question. Ahmed is best helped by being guided to solve this himself rather than just giving him the answer. \$\endgroup\$ – Graham Nye Apr 6 '20 at 12:27

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