A capacitor of \(2.0 \times 10^{-11 F }\) and an inductor are joined in series. The value of the inductance that will give the circuit a resonant frequency of \(200 kHz\) is ____________
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option D
Explanation:
\begin{aligned}
&\text { At resonance } \\
&X_{L}=X_{C} \\
&2 \pi f L=\frac{1}{2 \pi \times f C} ; L=\frac{1}{2 \pi^{2} f^{2} C} \\
&L=\frac{1}{4 \times 3.142^{2} \times\left(2 \times 10^{2}\right)^{2} \times 2 \times 10^{-11}} \\
&L=0.031=\frac{1}{32} H
\end{aligned}
\begin{aligned}
&\text { At resonance } \\
&X_{L}=X_{C} \\
&2 \pi f L=\frac{1}{2 \pi \times f C} ; L=\frac{1}{2 \pi^{2} f^{2} C} \\
&L=\frac{1}{4 \times 3.142^{2} \times\left(2 \times 10^{2}\right)^{2} \times 2 \times 10^{-11}} \\
&L=0.031=\frac{1}{32} H
\end{aligned}