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In a series R-L-C circuit, the phase angle is 40.5 ∘, with the source voltage leading the current. The reactance of the capacitor is 405 Ω and the resistance of the resistor is 205 Ω . The average power delivered by the source is 155 W .

a.Find the reactance of the inductor.

b.Find the rms current.

c.Find the rms voltage of the source.

Answer :

Khoso123

Answer:

(a) XL=580.1 Ω

(b) [tex]I_{rms}=0.87A[/tex]

(c) [tex]V_{rms}=234.43V[/tex]

Explanation:

For part (a)

Since we have given angle α, R and Xc.When the problem says that the source voltage lags the current it means α extends into 4th quadrant and is angle is negative

So

[tex]tan\alpha =\frac{X_{L}-X_{C}}{R} \\X_{L}=R.tan\alpha+X_{C}\\ X_{L}=(205)tan(40.5^{o})+405\\X_{L}=580.1ohms[/tex]

XL=580.1 Ω

For part (b)

We are also given the average power So we can find Irms easily

So

[tex]P_{avg}=I_{rms}^{2}R\\I_{rms}=\sqrt{\frac{P_{avg}}{R} } \\I_{rms}=\sqrt{\frac{155}{205} } \\I_{rms}=0.87A[/tex]

For Part (c)

We can find the Vrms by multiplying current I with impedance Z

So

[tex]V_{rms}=I_{rms}Z\\V_{rms}=(0.87)\sqrt{(205)^{2}+(580.1-405)^{2}}\\V_{rms}=234.43V[/tex]

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