A circuit containing resistance R_1,inductanc | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For a series $RLC$ circuit,the resonant frequency is given by $\omega_r = \frac{1}{\sqrt{LC}}$.Given that both circuits have the same resonant fre

Vedclass · Accepted Answer

$f_r$

Vedclass · Answer

(D) For a series $RLC$ circuit,the resonant frequency is given by $\omega_r = \frac{1}{\sqrt{LC}}$.Given that both circuits have the same resonant frequency $f_r$,we have:$\omega_r = \frac{1}{\sqrt{L_1 C_1}} = \frac{1}{\sqrt{L_2 C_2}} \implies L_1 C_1 = L_2 C_2$.When the two circuits are connected in series,the equivalent inductance is $L_{eq} = L_1 + L_2$ and the equivalent capacitance is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$.The new resonant frequency $\omega'$ is given by:$\omega' = \frac{1}{\sqrt{L_{eq} C_{eq}}} = \frac{1}{\sqrt{(L_1 + L_2) \cdot \frac{C_1 C_2}{C_1 + C_2}}}$Substituting $L_1 = \frac{L_2 C_2}{C_1}$:$\omega' = \frac{1}{\sqrt{(\frac{L_2 C_2}{C_1} + L_2) \cdot \frac{C_1 C_2}{C_1 + C_2}}} = \frac{1}{\sqrt{L_2 C_2 (\frac{C_2 + C_1}{C_1}) \cdot \frac{C_1 C_2}{C_1 + C_2}}} = \frac{1}{\sqrt{L_2 C_2}} = \omega_r$.Thus,the new resonant frequency remains $f_r$.

$A$ circuit containing resistance $R_1$,inductance $L_1$,and capacitance $C_1$ connected in series resonates at the same frequency $f_r$ as another circuit containing $R_2$,$L_2$,and $C_2$ in series. If the two circuits are connected in series,then the new frequency at resonance is

Similar Questions

In the circuit shown in the figure,an a.c. source gives voltage $V = 20 \cos (2000 t)$. The impedance and r.m.s. current respectively will be:

For a series $LCR$ circuit at resonance,the statement which is not true is:

$A$ resistor of $100 \Omega$, an inductor of $\frac{25}{\pi^2} \text{ mH}$ and a capacitor of $0.1 \mu\text{F}$ are connected in series to an $AC$ source. The impedance of the circuit is minimum for a frequency of (in $\text{ kHz}$)

The reading of the ammeter in the circuit shown will be: (in $A$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module