Question

A single loop consists of inductors $\mathbf{(}{\mathit{L}}_{\mathbf{1}}\mathbf{,}{\mathit{L}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{)}$, capacitors $\mathbf{(}{\mathit{C}}_{\mathbf{1}}\mathbf{,}{\mathit{C}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{)}$, and resistors $\mathbf{(}{\mathit{R}}_{\mathbf{1}}\mathbf{,}{\mathit{R}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{)}$ connected in series as shown, for example, in Figure-a. Show that regardless of the sequence of these circuit elements in the loop, the behavior of this circuit is identical to that of the simple LCcircuit shown in Figure-b. (Hint:Consider the loop rule and see problem) Problem:- Inductors in series.Two inductors L₁ and L₂ are connected in series and are separated by a large distance so that the magnetic field of one cannot affect the other.(a)Show that regardless of the sequence of these circuit elements in the loop, the behavior of this circuit is identical to that of the simple LC circuit shown in above figure (b). (Hint: Consider the loop rule)

Short answer

The behavior of the circuit in figure (a) is identical to that simple LC circuit shown in figure (b) regardless of the sequence of these circuit elements.

Step-by-step solution

The given data

Inductors $\left(L_1,L_2,\dots \dots \right)$, Capacitors $\left(C_1,C_2,\dots \dots \right)$, Resistors$\left(R_1,R_2,\dots \dots \right)$ are connected in a series.

Understanding the concept of Kirchhoff’s loop rule

Kirchhoff's loop rule states that the sum of all the electric potential differences around a loop is zero. By using the loop rule and differential equation for damped oscillations in the RLC circuit, we will prove that the behaviour of LC circuit (a) is identical to that of (b).

Formulae:

The emf equation, according to Kirchhoff’s loop rule,${\epsilon }_{total}=0$ (i)

Calculation of the behaviors of both the circuits

Applying the loop rule to circuit (a), we get that the total emf of the circuit is given by,

$\begin{array}{rcl}{\epsilon }_{total}& =& {\epsilon }_{L1}+{\epsilon }_{C1}+{\epsilon }_{L2}+{\epsilon }_{R1}+{\epsilon }_{C2}+{\epsilon }_{R2}+\dots \dots \dots \dots .\\ & =& 0\end{array}$
Let$j=1,2,3,\dots ..$

Then, the total emf can be written as:

${\epsilon }_{total}=\sum _j\left({\epsilon }_{Lj}+{\epsilon }_{Cj}+{\epsilon }_{Rj}\right)$

By using the differential of the above equation for the LCR circuit, we get that ,

$\frac{d{\epsilon }_{total}}{dt}=\sum _j\left(L_j\frac{di}{dt}+\frac{q}{C_j}+i{R}_j\right)$

$\begin{array}{rcl}\sum _j{L}_j& =& L,\\ \sum _j{C}_j& =& C,\\ \sum _j{R}_j& =& R\\ & & \end{array}$

$L\frac{di}{dt}+\frac{q}{C}+iR=0\text{(}\because \text{from equation (i))}$

This is equivalent to the simple LRC circuit (b).

Hence circuit (a) is identical to the circuit (b).