When a resistor is connected across the terminals of an acgenerator (104 V) that
ID: 1672522 • Letter: W
Question
When a resistor is connected across the terminals of an acgenerator (104 V) that has a fixedfrequency, there is a current of 0.505 Ain the resistor. When an inductor is connected across the terminalsof this same generator, there is a current of 0.384 A in the inductor. When both the resistor andthe inductor are connected in series between the terminals of thisgenerator, calculate the following. (a) the impedance of the series combination1
(b) the phase angle between the current and the voltage of thegenerator
2°
thanks for your help! :)
(a) the impedance of the series combination
1
(b) the phase angle between the current and the voltage of thegenerator
2°
thanks for your help! :)
Explanation / Answer
a. Resistance R = V/ IR = 104/ 0.505 = 205.94 Inductivereactance XC = V/ IC = 104/ 0.384 = 270.83 When both inductor andresistor are connected net impedance Z = (R2 + XL2) = (205.942 + 270.832) = 340.24 b. Phaseangle = tan-1(XL / R) = tan-1(270.83 / 205.94) = 52.750 = 104/ 0.384 = 270.83 When both inductor andresistor are connected net impedance Z = (R2 + XL2) = (205.942 + 270.832) = 340.24 b. Phaseangle = tan-1(XL / R) = tan-1(270.83 / 205.94) = 52.750Related Questions
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