contestada

The AC voltage source supplies an rms voltage of 146 V at frequency f. The circuit has R = 110 Ω, XL = 210 Ω, and XC = 110 Ω. At the instant the voltage across the generator is at its maximum value, what is the magnitude of the current in the circuit?

Respuesta :

Answer:

1.03A

Explanation:

For computing the magnitude of the current in the circuit we need to do the following calculations

LCR circuit impedance

[tex]Z = \sqrt{R^2 + (X_L - X_c)^2} \\\\ = \sqrt{110^2 + (210 - 110)^2}[/tex]

= 148.7Ω

Now the phase angle is

[tex]\phi = tan^{-1} (\frac{X_L - X_C}{R}) \\\\ = tan^{-1} (\frac{210 - 110}{110})\\\\ = 42.3^{\circ}[/tex]

Now the rms current flowing in the circuit is

[tex]I_{rms} = \frac{V_{rms}}{Z} \\\\ = \frac{146}{148.7}[/tex]

= 0.98 A

The current flowing in the circuit is

[tex]I = I_{rms}\sqrt{2} \\\\ = (0.98) (1.414)[/tex]

= 1.39 A

And, finally, the current across the generator is

[tex]I'= I cos \phi[/tex]

[tex]= (1.39) cos 42.3^{\circ}[/tex]

= 1.03A

Hence, the magnitude of the circuit current is 1.03A

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