When only a resistor is connected across the terminals of an ac generator (103 V

Question

When only a resistor is connected across the terminals of an ac generator (103 V) that has a fixed frequency, there is a current of 0.490 A in the resistor. When only an inductor is connected across the terminals of this same generator, there is a current of 0.412 A in the inductor. When both the resistor and the inductor are connected in series between the terminals of this generator, calculate the following.

(a) the impedance of the series combination
_____________ ?

(b) the phase angle between the current and the voltage of the generator
_____________

Explanation / Answer

apply Ohms law as V = iR

where i is current

R is resistance

V is Volatge

so

Resisatnce R = V/i = 103/0.490 = 210.2 ohms
---------------------

inductive reactance XL = V/i = 103./0.412 = 250 ohms

so

now use the formula for Impedence

Z^2 = R^2+ *Xl^2

so

Z^2 = 210.2^2 +(250^2)

Z = 326.625 ohms

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

apply phase angle tan theta = (XL/R)

tan theta = 250/210.20

tan theta = 1.189

theta = 49.94 deg

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