Question

A circuit containing a $80mH$ inductor and $60\mu F$ capacitor in series is connected to a $230V,50Hz$ supply. If the resistance in the circuit is negligible then the current amplitude will be

• A. 11.63 A
• B. 8.23 A
• C. 9.2 A
• D. 13.67 A

Answer 1

Answer: (B)

8.23 A

$\omega =2\pi f=100\pi$ ; $f=50Hz$

Impedance of circuit $Z=j(L\omega -\frac{1}{C\omega })=j(80\times {10}^{-3}\times 100\pi -\frac{1}{60\times {10}^{-6}\times 100\pi })$

Current amplitude $=\left|\frac{V}{Z}\right|=\frac{230}{|L\omega -\frac{1}{C\omega }|}=8.226A(\because \omega =2\pi 50)$