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  • A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having dampling constant 'b', the correct equivalence would be:-   Option: 1

A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having dampling constant 'b', the correct equivalence would be:-
 

Option: 1 L\leftrightarrow k,C\leftrightarrow b,R\leftrightarrow m


Option: 2 L\leftrightarrow m,C\leftrightarrow k,R\leftrightarrow b


Option: 3 L\leftrightarrow m,C\leftrightarrow \frac{1}{k},R\leftrightarrow b  


Option: 4 L\leftrightarrow \frac{1}{b},C\leftrightarrow \frac{1}{m},R\leftrightarrow \frac{1}{k}
 

Answers (1)

best_answer

 

 

Series LCR circuit -

Series LCR circuit-

                                  

The Figure given above shows a circuit containing a capacitor ,resistor and inductor connected in series through an alternating / sinosoidal voltage source.

As they are in series so the same amount of current will flow in all the three circuit components and for the voltage, vector sum of potential drop across each component would be equal to the applied voltage.

Let 'i' be the amount of current in the circuit at any time and VL,VC and VR the potential drop across L,C and R respectively then
                                          \begin{array}{l}{\mathrm{v}_{\mathrm{R}}=\mathrm{i} \mathrm{R} \rightarrow \text { Voltage is in phase with i }} \\ \\ {\mathrm{v}_{\mathrm{L}=\mathrm{i} \omega \mathrm{L}} \rightarrow \text { voltage is leading i by } 90^{\circ}} \\ \\ {\mathrm{v}_{\mathrm{c}}=\mathrm{i} / \mathrm{\omega} \mathrm{c} \rightarrow \text { voltage is lagging behind i by } 90^{\circ}}\end{array}\varepsilon

By all these we can draw phasor diagram as shown below -

                                

One thing should be noticed that we have assumed that VL is greater than VC which makes i lags behind V. If VC > VL then i lead V. So as per our assumption, there resultant will be (VL -VC). So, from the above phasor diagram V will represent resultant of vectors VR and (VL -VC). So the equation become - 

                                                   \begin{aligned} V &=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}} \\ \\ &=i \sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \\ \\ &=i \sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}} \\ \\ &=i Z \\ \text { where, } & \\ Z &=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}} \end{aligned}

Here, Z is called Impedence of this circuit.

Now come to the phase angle. The phase angle for this case is given as - 

                                            \tan \varphi=\frac{V_{L}-V_{C}}{V_{R}}=\frac{X_{L}-X_{C}}{R}=\frac{\omega L-\frac{1}{\omega C}}{R}

Now from the equation of the phase angle three cases will arise. These three cases are - 

(i) When,     \omega L \ > \ \frac{1}{\omega C}

    then, tanφ is positive i.e. φ is positive and voltage leads the current i.
(ii) When \omega L \ < \ \frac{1}{\omega C}

    then, tanφ is negative i.e. φ is negative and voltage lags behind the current i.
(iii) When  \omega L \ = \ \frac{1}{\omega C} ,

      then tanφ is zero i.e. φ is zero and voltage and current are in phase. This is called electrical resonance. 

 

 

-

 

 

For the spring-mass damping system, the governing equation is given by

m \frac{dx^2}{dt^2}+b\frac{dx}{dt}+kx=0...............(1)

For an LCR damped oscillator, the equation is 

L \frac{dQ^2}{dt^2}+R\frac{dQ}{dt}+\frac{Q}{C}=0...............(2)

Comparing 1 and 2

\\L\leftrightarrow m\\C\leftrightarrow \frac{1}{k}\\R\leftrightarrow b

So the answer is option (3)

Posted by

Ritika Jonwal

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