Question

An alternating voltage E = 6 sin 20t + 8 cos 20t is applied to a series resonant circuit as shown. The correct statements are –

• A. The capacitance C is 12.5 Mf
• B. The resonant current in the circuit is 3A
• C. Power dissipated in the circuit is 20 watt
• D. Quality factor of the current is 0.9

Answer 1

Answer: (A)

The capacitance C is 12.5 Mf

Answer 2

E = 10 sin (20t + f)

tan f = $\frac{8}{6}$

i = $\frac{10}{\sqrt{2R}}$= $\frac{10}{R}$ = $\frac{2amp}{\sqrt{2}}$

Power dissipated = i²R = $\frac{4 \times 5}{2}$

= 10 watt

\ C = $\frac{1}{\omega^{2}L}$= $\frac{1}{400 \times 0.2}$

= 12.5 mF