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An AC voltage source of amplitude 10V is supplied across an inductor and a resistor wired in series. The voltage amplitude across the inductor is 5V. The voltage amplitude across the resistor is closest to:___________.

Respuesta :

Answer:

8.66 V

Explanation:

Vo = 10 v

VL = 5 V

VR = ?

According to the equation of LR circiut

[tex]V_{0}^{2}=V_{R}^{2}+ V_{L}^{2}[/tex]

10 x 10 = VR² + 5 x 5

100 - 25 = VR²

VR² = 75

VR = 8.66 V

Thus, the voltage amplitude across the resistor is 8.66 v.

Explanation:

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