Question

In the critically damped circuit shown in the figure below, the initial conditions on the storage elements are i_L(0) = 2 A and v_C(0) = 5 V. Determine the voltage v(50 ms).

In the critically damped circuit shown in the figure below, the initial conditions on the storage elements are i(0) = 2 A and vc0) = 5 V. Determine the voltage v(50 ms). + + il(0) vc( 0+ 0.01 F v(t) 3 1002 4H X 9.72 V the tolerance is +/-2%

Please show all work, thank you.

Answer 1

+ + it( υc (0) 0.01E υ(1) 10 Ω 4 Η

Writing KCL equation at Node A:

dVe V 10 + so Vedt + I (0) + 0.014 C=0.... dt .(0)

NOTE: Voltage across all the elements are same and we have to consider the initial current "Io" while writing the KCL equation.

Differentiate Equation (i):

$\frac{1}{100}\frac{d^2V_c}{dt^2}+\frac{1}{10}\frac{dV_c}{dt}+\frac{1}{4}V_c=0$

$\frac{d^2V_c}{dt^2}+{10}\frac{dV_c}{dt}+25V_c=0\dots\dots\dots\dots\dots(ii)$

In the Laplace Domain:

(s? + 10s + 25)V.=0

$V_c\neq 0$

$(s^2+10s+25)=0$

So, the solution is:

$s=-5$

So, the solution to the differential equation is:

$V_c(t)=Ae^{st}+Bte^{st}$

$V_c(t)=A\cdot e^{-5t}+B\cdot t\cdot e^{-5t}\dots\dots\dots\dots\dots(iii)$

Now from the initial Conditions, find A and B:

$\left |V_c(t) \right |_{t=0}=\left |A\cdot e^{-5t}+B\cdot t\cdot e^{-5t} \right |_{t=0}$

$0=A\cdot e^{0}+B\cdot t\cdot e^{0}$

$A+B\cdot0=0$

$A=0\dots\dots\dots\dots\dots(iv)$

From equation (i):

$\left |\frac{V_c}{10}+I_L(0)+0.01\frac{dV_c}{dt} \right |_{t=0}=0$

[Integral term can be omitted as the integration was not started at t=0]

$\left |\frac{dV_c}{dt} \right |_{t=0}=-\frac{5}{10}-2=-2.5$

Differentiating eq (iii):

$\frac{dV_c(t)}{dt}=-5A\cdot e^{-5t}-5B\cdot t\cdot e^{-5t}+Be^{-5t}$

$\left |-5A\cdot e^{-5t}-5B\cdot t\cdot e^{-5t}+Be^{-5t} \right |_{t=0}=-2.5$

$-5A+B=-2.5$

Put the value of A from eq (iv)

$B=-2.5$

So, the equation for capacitor Voltage becomes:

$V_c(t)=V_c(0)-2.5te^{-5t}$

The voltage at 50 ms:

$V_c(t)=5-2.5\times50\times10^{-3}e^{-5\times50\times10^{-3}}$

$V_c(t)\approx 4.9\ V$