All of the questions in this problem are based on the circuit below. R=10Ω,L=5mH, C=500μF. The source voltage is 10cos(200t+45LaTeX: ^{^\circ} ∘ ). Round all of your answers to two decimal places if necessary. Omit the units. What is the inductor impedance value in ohms? First, what is the REAL part.

1) 0 Ω

2) jΩ

3) 0 Ω

4) -10j Ω

5)  10Ω

6) -9j Ω

Explanation:

Given that R=10Ω,L=5mH,C=500μF and The source voltage is 10cos(200t+45°)

The source voltage is given by

Therefore comparing with the source voltage, the angular frequency w = 200 rad/s

a) The impedance of the inductor is given by:

The real part is 0 Ω and

the imaginary part is j Ω

b) The impedance of the capacitor is given by:

The real part is 0 Ω and

the imaginary part is -10j Ω

c) The total impedance of the circuit is given by:

The real part is 10Ω

The Imaginary part is -9j Ω

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation:

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