Question

(1 point) In an RLC series circuit, the source has a peak potential difference of 120 V, and a frequency of 55 Hz. Take R=1212, C = 170 uF, L = 0.3 H. Find: a) X and Xc; XL XC = 12 = 12 b) the phase angle; degrees c) the average power delivered by the source; W d) the power factor.

Answer 1

Note that power factor value is 0.505

Power factor is always less than unity and it is unity only in the case of resonance.

We know that Xe = inductive reactance WL QTfL where f = frequency. L = inductor value. X e = e = capacitive reactance wc di fc / we get, XL = 2x11x55 * 0.3=1332 T xe = alfc 2*T* 55 x 170*108 Xc = 53.47 2 site

(b) phase angle is given by. tano= Xe - Xe R. XL - Xc R 33 - 53.47 12 capacitor dominating can'( 59.60 negative sign implies it is circuit (c) Power delivered will only be due to R as l'anda Stores energy and are responsible in phase chance. hence Pow=v² R. Power: 1200 W (d, power factor is nothing but cos o where z = impedence given by = √ R² + (x2-xcj? hence 2 = 23.72 2 We get Caso 0.505 23.72
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