Answer:
(A) 2652.49 ohm (b) 91937.45311 Hz (c) (i) 12.022 A (II) 2.324 A
Explanation:
We have given resistance R = 10 ohm
Capacitance C = 1 nF
Inductance of the coil L = 3 mH
(A) Inductive reactance [tex]X_L=\omega L=377\times 3\times 10^{-3}=1.131ohm[/tex]
Capacitive reactance [tex]X_C=\frac{1}{\omega C}=\frac{1}{377\times 10^{-9}}=2.6525\times 10^6ohm[/tex]
Impedance [tex]Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{10^2+(2652500-1.131)^2}=2652.49ohm[/tex]
(b) We know that resonance frequency [tex]f=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi \sqrt{3\times 10^{-3}\times 10^{-9}}}=91937.45311Hz[/tex]
(c) (i) At resonance condition [tex]X_L=X_C[/tex] so only effective resistance is R
So maximum current [tex]i=\frac{V}{R}=\frac{\frac{170}{\sqrt{2}}}{10}=12.022A[/tex]
(ii) Current across the coil [tex]i=\frac{voltage\ across\ the\ coil}{impedence\ of\ the\ coil}=\frac{\frac{3}{\sqrt{2}}}{1.131}=2.324A[/tex]
