Question

L, C and R are connected in series to an A.C. voltage V = Vm cosωt. Obtain the differential equation for the charge.

Answer 1

Answer:

$dq\ =\ \dfrac{Vm.cos\omega t}{\sqrt{R^2+j(\omega.L-\dfrac{1}{\omega.C})^2}}dt$

Explanation:

Given,

Voltage, V= Vm.cosωt

Since, the L, C, R are connected in series. Hence the total impedance connected in the circuit is

 $Z = \sqrt{R^2+j(\omega.L-\dfrac{1}{\omega.C})^2}$

Current is given by,

$I = \dfrac{V}{Z}$

$=\dfrac{Vm.cos\omega t}{\sqrt{R^2+j(\omega.L-\dfrac{1}{\omega.C})^2}}$

$but\ I\ =\ \dfrac{dq}{dt}$

      $\dfrac{Vm.cos\omega t}{\sqrt{R^2+j(\omega.L-\dfrac{1}{\omega.C})^2}}\ =\ \dfrac{dq}{dt}$

$dq\ =\ \dfrac{Vm.cos\omega t}{\sqrt{R^2+j(\omega.L-\dfrac{1}{\omega.C})^2}}dt$

Hence the differential equation of charge will be

$dq\ =\ \dfrac{Vm.cos\omega t}{\sqrt{R^2+j(\omega.L-\dfrac{1}{\omega.C})^2}}dt$

         

Answer 2

Answer:

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