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When a capacitor of small capacitance is connected in series with series L-R circuit. The alternating current in the circuit increases. Explain why? |
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Answer» Solution :Addition of capacitor in the given circuit decreases the impedance Z of the circuit and hence increases current I in the circuit as `I=(V)/(Z)" where "Z=sqrt(R^(2)+X_(L)^(2))" WITHOUT capacitor"` `"and new "Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))` with capacitor |
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