The figure shows a combination of inductances | vedclass

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

Question

(D) In the given circuit,the two inductors $L$ and $2L$ are connected in series. Therefore,the equivalent inductance $L_{eq}$ is given by:$L_{eq} = L

Vedclass · Accepted Answer

$\frac{1}{6 \pi \sqrt{LC}}$

Vedclass · Answer

(D) In the given circuit,the two inductors $L$ and $2L$ are connected in series. Therefore,the equivalent inductance $L_{eq}$ is given by:$L_{eq} = L + 2L = 3L$The two capacitors $C$ and $2C$ are connected in parallel. Therefore,the equivalent capacitance $C_{eq}$ is given by:$C_{eq} = C + 2C = 3C$The resonant frequency $f$ of an $L-C$ circuit is given by the formula:$f = \frac{1}{2 \pi \sqrt{L_{eq} C_{eq}}}$Substituting the values of $L_{eq}$ and $C_{eq}$ into the formula:$f = \frac{1}{2 \pi \sqrt{(3L)(3C)}}$$f = \frac{1}{2 \pi \sqrt{9LC}}$$f = \frac{1}{2 \pi \cdot 3 \sqrt{LC}}$$f = \frac{1}{6 \pi \sqrt{LC}}$

The figure shows a combination of inductances and capacitances. The resonant frequency of the $L-C$ circuit is:

Similar Questions

For a series $LCR$ circuit,$R = X_L = 2X_C$. The impedance of the circuit and the phase difference between $V$ and $I$ respectively will be:

In the adjoining figure,the impedance of the circuit will be $... \Omega$.

The angular frequency of alternating current in an $L-C-R$ circuit is $100 \, rad/s$. The components connected are shown in the figure. Find the value of the inductance of the coil and the capacity of the condenser.

Draw phasor diagrams for $X_C > X_L$ and $X_C < X_L$,and state the disadvantages of the phasor method.

Vedclass Test Series

Exam Paper Generator

Online Exam Module