Question

3. A 10 ohm resistance, 5 mH coil and 10 µF
capacitor are joined in series. When a suitable
frequency alternating current source is joined
to this combination, the circuit resonates. If
the resistance is halved, the resonance
frequency
(A) is halved.
(B) is doubled.
(C) remains unchanged.
(D) is quadrupled
​

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Resistance},\mathrm{R}=10\mathrm{ohm} \\ \mathrm{inductance},\mathrm{L}=10\mathrm{mH}=10\times {10}^{-3}\mathrm{H} \\ \mathrm{capacitance},\mathrm{C}=10\mathrm{\mu F}=10\times {10}^{-6}\mathrm{F} \\ \mathrm{Resonating}\mathrm{frequency}=\frac{1}{2\mathrm{\pi }\sqrt{\mathrm{LC}}} \\ \mathrm{Resonating}\mathrm{frequency}=\frac{1}{2\times 3.14\sqrt{10\times {10}^{-3}\times 10\times {10}^{-6}}} \\ \mathrm{Resonating}\mathrm{frequency}=\frac{1592.35}{\sqrt{10}}=503.9\mathrm{Hz} \\ \mathrm{From}\mathrm{the}\mathrm{above}\mathrm{formula},\mathrm{we}\mathrm{conclude}\mathrm{that}\mathrm{resonating}\mathrm{frequency}\mathrm{does}\mathrm{not}\mathrm{depend}\mathrm{on}\mathrm{resistance}. \\ \mathrm{Hence},\mathrm{it}\mathrm{will}\mathrm{remain}\mathrm{unchanged}. \end{gathered}$$