(22) A resistance of 3 ohms, a coil of inductance 0.01 H are
connected in series with a capacitor, and put across a 200 volt,
50 Hz supply. Calculate:
(i) the capacitance of the capacitor so that the circuit resonates.
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Answer:
at resonance we can say Xc=XL
where Xl=l omega=0.01*2pie*50=1/c*omega
=1/4pie square*2500*0.01=0.001 f
i took pie square as 10
Explanation:
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