When a resistor is connected across the terminals of an ac generator (120 V) tha

Question

When a resistor is connected across the terminals of an ac generator (120 V) that has a fixed frequency, there is a current of 0.580 A in the resistor. When an inductor is connected across the terminals of the same generator, there is a current of 0.639 A in the inductor. When both the resistor and the inductor are connected in series between the terminals of this generator, what is (a) the impedance of the series combination and (b) the phase angle between the current and the voltage of the generator? Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise.

Explanation / Answer

First we find Resistance value when there is only resistance connected in circuit using relation

V = IR so R = V/I = 120V/.580 A = 206.89 ohm

Similarly we find value of Inductor when there is only inductor in circuit using relation

XL = V/I = 120/.639 = 187.79 ohm

Imedance of series combination is given by Z = sqrt ( R2 + XL2 ) = sqrt ( 206.892 + 187.792 ) = 279.40 ohm

(i) Z = 279.40 Ohm

(ii) Phase angle is given by = arctan(XL/R) = arctan(187.79/206.89) = 42.22 degree

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