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## Answered question

2022-01-09

Analyzing an L-R-C Circuit. You have a 200 Ω resistor, a 0.400 H inductor, a 5.00 uF capacitor, and a variable-frequency ac source with an amplitude of 3.00 V. You connect all four elements together to form a series circuit. (a) At what frequency will the current in the circuit be greatest? What will be the current amplitude at this frequency? (b) What will be the current amplitude at an angular frequency of 400 rad/s? At this frequency, will the source voltage lead or lag the current?

### Answer & Explanation

Shawn Kim

Beginner2022-01-10Added 25 answers

Given:
The resistance is 200 ohms.
The inductance is 0.400 H.
The capacitance is 5.00 μF.
The voltage of the source is 3.00 V.
The angular frequency is 400 rad/s.
a) Calculation:
Write the expression for the frequency
${f}_{o}=\frac{1}{2\pi \sqrt{LC}}$
Substitute the values.

Write the expression for the ohms

jean2098

Beginner2022-01-11Added 38 answers

b) Calculation:
Write the expression for the impedance.
$Z=\sqrt{{R}^{2}+\left(2\pi fL-\frac{1}{2\pi fC}{\right)}^{2}}$
Substitute the values.
$Z=\sqrt{{\left(200\right)}^{2}+\left[\left(400\right)\left(0.4\right)+{\frac{1}{\left(400\left(5.00×{10}^{-6}\right)\right\}}}^{2}\right\}}$
$=394.5$
Calculate the current.
$I=\frac{V}{Z}$
Substitute the values.

Calculate the phase difference
$\varphi ={\mathrm{tan}}^{-1}\frac{\omega L-\omega C}{R}$
Substitute the values.
$\varphi ={\mathrm{tan}}^{-1}\left[\frac{\left(400\right)\left(0.4\right)-\frac{1}{\left(400\left(5.00×{10}^{-6}\right)\right\}}\left\{200\right\}}{}$
$=-{59.53}^{\circ }$
Thus, the current is 7.6 mA and the voltage lags behind the current by 59.53 degress.

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