When only a resistor is connected across the terminals of an ac generator (108 V

Question

When only a resistor is connected across the terminals of an ac generator (108 V) that has a fixed frequency, there is a current of 0.505 A in the resistor. When only an inductor is connected across the terminals of this same generator, there is a current of 0.432 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

Remember Capacitors work like parallel branches and inductuctors act like resistors in series. Another mnemonic low freq passing represents an inductive branch, while high frequency passing represents a capacitive branch.
With ac circuits you construct a triangle with impedance as the hypotenuse and resistance r on the base and reactance x on the altitude.
R=108/0.505 = 213.86 ohms
Ri=108/0.432 =250;

z=sqrt(213.86^2+250^2)=329 ohms

b. phase angle is 90 degrees between current and voltage

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