Question

A condenser of $250\mu F$ is connected in parallel to a cell of inductance 0.16 mH, while its effective resistance is $20 \Omega$. Determine the resonant frequency.
A. $9 \times {10}^{4} Hz$
B. $16 \times {10}^{7} Hz$
C. $8 \times {10}^{5} Hz$
D. $9 \times {10}^{3} Hz$

Answer 1

Hint: In the question it is mentioned that the capacitor is connected in parallel to inductor and the values of resistance are given. So, the circuit is a parallel LCR circuit. To solve this problem, use the formula for resonant frequency of parallel LCR series. Substitute the values in the formula and find the value of resonant frequency.

Formula used:
$f= \dfrac {1}{2 \pi} \sqrt {\dfrac {1}{LC}- \dfrac {{R}^{2}}{{L}^{2}}}$

Complete step by step answer:
Given: Capacitance (C)= $250\mu F= 250 \times {10}^{-6}F$
              Inductance (L)= 0.16 mH=$ 0.16 \ times {10}^{-3} H$
              Resistance (R)= $20 \Omega$
We know, resonant frequency of a parallel LCR series is given by,
$f= \dfrac {1}{2 \pi} \sqrt {\dfrac {1}{LC}- \dfrac {{R}^{2}}{{L}^{2}}}$
Substituting values in above equation we get,
$f= \dfrac {1}{2 \pi} \sqrt {\dfrac {1}{250 \times {10}^{-6} \times 0.16 \times {10}^{-3}}- \dfrac {{20}^{2}}{({0.16 \times {10}^{-3})}^{2}}}$
$\Rightarrow f= \dfrac {1}{2 \pi} \sqrt {\dfrac {1}{40 \times {10}^{-9}}- \dfrac {400}{2.56 \times {10}^{-8}}}$
$\Rightarrow f=\dfrac{1}{2\pi} \times 50.14 \times {10}^{5}$
$\Rightarrow f= 7.98 \times {10}^{5} Hz$
$\therefore f \approx 8 \times {10}^{5}$
Hence, the resonant frequency is $8 \times {10}^{5}$.
So, the correct answer is option C i.e. $8 \times {10}^{5}$.

Note:
Students must read the question carefully and see whether the question is related to parallel LCR circuit or series LCR circuit. According to the type of circuit select the formula for resonant frequency. If the question was related to series LCR circuit, then the formula for resonant frequency used will be given by,
$f= \dfrac {1}{2 \pi \sqrt {LC}}$.