An inductor (L=530mH), a capacitor (C=6.50?F(micro farad), and a resistor (R=660

Question

An inductor (L=530mH), a capacitor (C=6.50?F(micro farad), and a resistor (R=660?(ohms) are connected in series. A 38.0Hz ac generator produces a peak current of 330mA in the circuit.

a) What is the impedance of the circuit? Answer in units of ?(ohms).

b) What is the required maximum voltage?

c) What is the phase angle by which current leads or lags the applied voltage? Answer in degrees. Be sure to include the sign to indicate whether the current leads (negative) or lags (positive) the voltage.

d) To what frequency should the ac generator be tuned to obtain the maximum current? Answer in Hz.

Explanation / Answer

Impedence Z^2 = R^2 + (XL-Xc)^2

XL = wL = 2pifL = 2*3.14 * 38 * 0.530 = 128.38 ohms

Xc = 1/WL = 1/(2*3.14* 38 * 6.5 e-6)   = 644.67 ohms

so

part A: impedence Z^2 = 660^2+(644.67-128.38)^2

Z = 838 ohms

-----------------------------

as Current I = V/Z

Vpeak = IZ

Vpeak = 0.33 * 838

Vpeak = 276.52 Volts

---------------------------------

phase angle tan theta   = (XL-Xc)/.R

tan theta   = (128.38 - 644.67)/660

tan theta = -0.7822

theta = -38.03 deg   ( leads)

--------------------------------

as givren currrnt is max , frequerncy f = 38 Hz

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