Question

A circuit containing Inductance L capacitance C in senes with applied electromotive force E. By Kirchhoffs voltage aw differential equation for
current is​

Answer 1

Answer:

The RL circuit shown above has a resistor and an inductor connected in series. A constant voltage V is applied when the switch is closed.

The (variable) voltage across the resistor is given by:

\displaystyle{V}_{{R}}={i}{R}V  

R

​  

=iR

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RL circuit examples

Two-mesh circuits

The (variable) voltage across the inductor is given by:

\displaystyle{V}_{{L}}={L}\frac{{{d}{i}}}{{{\left.{d}{t}\right.}}}V  

L

​  

=L  

dt

di

​  

 

Kirchhoff's voltage law says that the directed sum of the voltages around a circuit must be zero. This results in the following differential equation:

\displaystyle{R}{i}+{L}\frac{{{d}{i}}}{{{\left.{d}{t}\right.}}}={V}Ri+L  

dt

di

​  

=V

Step-by-step explanation:

Answer 2

Answer:

The second order differential equation of LC circuit for current is$\frac{d^{2} I}{dt} +\frac{1}{LC} I=0$

Step-by-step explanation:

Given a circuit containing Inductance L, capacitance C  and a source of  electromotive force E in series.

Applying Kirchhoff's voltage law to the LC circuit, gives

$V_{L} +V_{C}=0$                                                              ...(1)

Using the constituent current-voltage relations for the inductor and capacitor,

$V_{L} =L\frac{dI}{dt}$  and $V_{C} =\frac{1}{C} \int Idt$ in equation (1),

$L\frac{dI}{dt} +\frac{1}{C} \int Idt=0$

Differentiating with respect to time, t

$\frac{d}{dt} \Big( L\frac{dI}{dt} +\frac{1}{C} \int Idt\Big)=0$

$\implies L\frac{d^{2} I}{dt} +\frac{1}{C} I=0$

Dividing the equation with L,

$\implies \frac{d^{2} I}{dt} +\frac{1}{LC} I=0$

This equation is the second order differential equation of LC circuit for current.