Question

A coil having a resistance and inductance of 5-ohm and 32 mH respectively is connected in series with 796pF

capacitor. Determine the resonant frequency of the circuit.​

Answer 1

Answer:

$\boxed{\mathfrak{Resonant \ frequency \ (f) = 3.2 \times 10^4 \ Hz}}$

Given:

Resistance = 5 Ω

Inductance (L) = 32 mH = $\sf 32 \times 10^{-3}$ H

Capacitance (C) = 796 pF = $\sf 796 \times 10^{-12}$ F

Explanation:

Resonant frequency (f) formula:

$\boxed{ \bold{f = \dfrac{1}{2 \pi \sqrt{LC}}}}$

By substituting values in the formula we get:

$\rm \implies f = \dfrac{1}{2\pi \sqrt{32 \times {10}^{ - 3} \times 796 \times {10}^{ - 12} } } \\ \\ \rm \implies f = \dfrac{1}{2 \times 3.14 \times \sqrt{2547.2 \times {10}^{ - 14} } } \\ \\ \rm \implies f = \frac{1}{2 \times 3.14 \times 50.47 \times {10}^{ - 7} } \\ \\ \rm \implies f = 3.2 \times {10}^{4} \: Hz$

Answer 2

Answer:

$\blue{\boxed{\tt{Resonant \ frequency \ (f) = 3.2 \times 10^4 \ Hz}}}</p><p>$

$\green{ \sf 32 \times 10^{-3}}$

Given:

Resistance = 5 Ω

Inductance (L) = 32 mH =

Capacitance (C) = 796 pF =

$\red{\sf 796 \times 10^{-12}}</p><p>$

Explanation:

Resonant frequency (f) formula:

$\orange{\boxed{ \bold{f = \dfrac{1}{2 \pi \sqrt{LC}}}} }</p><p>$

By substituting values in the formula we get:

$\pink{\begin{gathered}\rm \implies f = \dfrac{1}{2\pi \sqrt{32 \times {10}^{ - 3} \times 796 \times {10}^{ - 12} } } \\ \\ \rm \implies f = \dfrac{1}{2 \times 3.14 \times \sqrt{2547.2 \times {10}^{ - 14} } } \\ \\ \rm \implies f = \frac{1}{2 \times 3.14 \times 50.47 \times {10}^{ - 7} } \\ \\ \rm \implies f = 3.2 \times {10}^{4} \: Hz\end{gathered}}</p><p>$