RLC Circuit - Power Dissipation Computation

$\textcolor{#20A900}{FunProblem}$

I will be happy if you continue this series.

The basic equations are, $q_{1}$ is the charge through resistor , $q_{2}$ is the charge through inductor. $10-\dot{q_1}-(q_1+q_2) =0$ $10-4\ddot{q_2}-(q_1+q_2) =0$ After solving this equations $\Large q_{1}(t) =c_{1}e^{\frac{-t}{2}}(e^{\frac{+t}{2}}-t)-c_{2}e^{\frac{-t}{2}}t-2c_{3}e^{\frac{-t}{2}}(-t+e^{\frac{t}{2}}-2)$ $\Large q_{2}(t) =\frac{-1}{2}c_{1}e^{\frac{-t}{2}}(-t+2e^{\frac{t}{2}}-2) +\frac{1}{2}c_{2}e^{\frac{-t}{2}}(t+2) +c_{3}e^{\frac{-t}{2}}(-t+4e^{\frac{t}{2}}-4)+10$ where $c_1, c_2, c_3$ are arbitrary constants. Use initial conditions $q_{1}(0) =0$ $q_{2}(0) =0$ $\dot{q_{1}}(0) =0$ $\dot{q_{2}}(0) =0$ After this $\large \textcolor{#BA33D6}{ \dot{q_{1}}(t) =I_{1}(t) =-5e^{\frac{-t}{2}}(t-2) }$ Using this $P=I^{2}R$ And substituting values and at $t=5$ $P=225e^{-5}$ $\textcolor{#3D99F6}{\boxed{a+b=220}}$

Let at any time $t$ , the current through the inductance be $I_1$ , the current drawn from the source be $I$ , and the charge on the capacitor be $q$ . Then

$I=\dfrac {dq}{dt}, R(I-I_1)=L\dfrac{dI_1}{dt}=E-\dfrac{q}{C}$

Substituting values we get

$4\dfrac{d^2I_1}{dt^2}+4\dfrac{dI_1}{dt}+I_1=0$

Solution to this equation is

$I_1=Ate^{-\frac{t}{2}}, I=A(4-t)e^{-\frac{t}{2}}$ , where $A$ is a constant of integration.

Solving for $q$ we get

$q=10-4A(1-\frac{t}{2})e^{-\frac{t}{2}}$ .

So, $A=\dfrac{5}{2}$ , and $P(t)=(I-I_1)^2R$

$=100(1-\frac{t}{2})^2e^{-t}$ .

At $t=5, P(t)=P(5)=100\times \dfrac {9}{4}\times e^{-5}=225e^{-5}$ .

So, $a=225,b=-5$ and $a+b=225-5=\boxed {220}$ .

The power dissipated in the resistor is $P(t) = \dfrac{v(t)^2}{R}$ . Using the Laplace transform, and by voltage divider,

$V(s) / E(s) = \displaystyle \dfrac{ \dfrac{ s LR }{sL + R} }{ \dfrac{1}{sC} + \dfrac{ s LR }{sL + R} }$

Multiplying top and bottom by $sC (sL + R)$ ,

$V(s)/E(s) = \displaystyle \dfrac{ s^2 LRC }{ s L + R + s^2 LRC }$

Now $E(s) = \dfrac{10}{s}$ , and substituting $L = 4, R = 1 , C = 1$ , we obtain,

$V(s) = \displaystyle \dfrac{ 40 s }{ 4 s^2 + 4 s + 1 } = \dfrac{40 s}{(2s+1)^2}$

Using partial fractions,

$\dfrac{40 s}{(2s+1)^2} = \dfrac{A}{2s + 1} + \dfrac{B}{(2s+1)^2}$

So that,

$40 s = A (2 s + 1) + B$

from which , $A = 20, B = -20$ . Therefore,

$V(s) = \dfrac{20}{2 s + 1} - \dfrac{20}{(2s+1)^2} = \dfrac{10}{s + \frac{1}{2} } - \dfrac{5}{(s + \frac{1}{2})^2}$

Applying the inverse Laplace transform,

$v(t) = 10 e^{-\frac{1}{2} t } - 5 t e^{-\frac{1}{2} t }$

At $t = 5$ ,

$v(5) = - 15 e^{-\frac{5}{2} }$

So that,

$P(5) = \dfrac{ v(5)^2}{1} = 225 e^{-5} = a e^b$

Hence, $a + b = 225 + (-5) = 220$