Question

A coil of inductance 0.4mH is connected to a capacitor 400 pf . To what length is this circuit tuned?

Answer 1

Resonance frequency is given as,

f= 1/2pie underroot LC

f=1/2×3.14 underroot (0.4×10power-3 ×400×10power-12)

=6.28×4×10power-7

=10power7/6.28×4

=3.98×10power5 Hz

Thus wavelength is given by,

λ=c/f

= (3×10power8 / 3.98×10power5)

=753.7m

HOPE U UNDERSTOOD

Answer 2

$\huge† \huge \bold{\: \pmb {\red{ Answer }} }$

Resonance frequency is given as,

$\sf \: f= \frac{1}{2\pi \sqrt{LC} } \\ \\ \sf \: f= \frac{1}{2×3.14 \sqrt{(0.4× {10}^{ - 3} ×400× {10}^{ - 12})}} \\ \\ \sf \: = \frac{1}{6.28×4× {10}^{ - 7} } \\ \\ \sf \: = \frac{ {10}^{7} }{6.28×4} \\ \\ \sf \red{ =3.98× {10}^{5} \: Hz}$

• Thus wavelength is given by,

$\sf \: λ= \frac{c}{f} \\ \\ \sf \: = \frac{3× {10}^{8} }{3.98× {10}^{5} } \\ \\ = \boxed{ \red{ \sf \: 753.7m}} \pink{\bigstar}$