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A pure resistance of 50 ohms is in series with a pure capacitance of 100 micro farads. The series combination is connected across 100-V, 50-Hz supply. Find (a) the impedance (b) current (c) power factor (d) phase angle (e) voltage across resistor (f) voltage across capacitor. Draw the vector diagram. |
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Answer» XC = 106 π /2π × 50 × 100 = 32 Ω ; R = 50 Ω (a) Z = √(502 + 322) + 59.4 Ω (b) I = V/Z = 100/59.4 = 1.684 A (c) p.f. = R/Z = 50 = 59.4 = 0.842 (lead) (d) φ = cos−1 (0.842)= 32°36′ (e) VR = IR = 50 × 1.684 = 84.2V (f) VC = IXC = 32 × 1.684 = 53.9V |
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