Topic: Network Theory
For the circuit shown in figure, find the frequency at which this circuit will be at resonance
- A. 1 rad/sec
- B. 2 rad/sec
- C. 0.25 rad/sec
- D. 0.5 rad/sec
Answer: D. 0.5 rad/sec
Explanation: $Z_{in} = j\omega + \frac{-j/2\omega}{1 - \frac{j}{2\omega}} = j\omega + \frac{-j}{2\omega-j} = j\omega + \frac{(-j)(2\omega+j)}{4\omega²+1}$
$Z_{in} = \frac{1+j(4\omega³-\omega)}{4\omega²+1}$
For resonance $Z_{in}$ should be real, so its imaginary part is zero.
$4\omega³-\omega=0 \Rightarrow \omega=0.5 \text{ rad/sec}$.