a 100 mH inductor , a 20 uF capacitor and a 10 ohm resistor are connected in series to a 100 V , 50 Hz ac source . calculate (i) impedance of the circuit at resonance (ii) current at resonance (iii) resonant frequency

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Solution

Dear Student,
$(3) A t r e s o n a n c e c o n d i t i o n w L = \frac{1}{w C} o r w^{2} = \frac{1}{L C} w = \frac{1}{\sqrt{L C}} s o r e s o n a n c e f r e q u e n c y = \frac{1}{\sqrt{. 1 \times 20 \times 10^{- 6}}} = 707 H z (1) a n d w e t o t a l i m p e n d e n c e = \sqrt{R^{2} + {(w L - \frac{1}{w C})}^{2}} i n c n d i t i o n o f r e s o n a n c e i m p e d e n c e = R (2) c u r r e n t = \frac{m a x i m u m v o l t a g e}{r e s i s \tan c e} = \frac{100}{10} = 10 A$
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a 100 mH inductor , a 20 uF capacitor and a 10 ohm resistor are connected in series to a 100 V , 50 Hz ac source . calculate (i) impedance of the circuit at resonance (ii) current at resonance (iii) resonant frequency

Dear Student, (3)At resonance condition wL=1wCor w2=1LCw=1LC so resonance frequency=1.1×20×10-6=707Hz(1)and we total impendence=R2+wL-1wC2in cndition of resonance impedence=R (2) current =maximum voltage resistance=10010=10A Regard