Determining an equation for the current after the switches close

Question

I know that the solution of of the form i(t) = A*sin(wt + phi), I am really confused where to start in finding A, w and phi however. Here is the circuit in question:

The switches close at time t=0

Answer 1

The initial conditions are \$\small v_C=2.5\:V\$ and \$\small i_L=\frac{5}{400}=12.5\:mA\$.

For \$\small t \ge 0\$, the same current, \$\small i\$, flows through \$\small L\$ and \$\small C\$. If this current is assumed to flow clockwise, KVL gives

$$\small 2.5-\frac{1}{C}\int i\:dt =L\frac{di}{dt}$$

Differentiating and rearranging gives the 2nd order ODE

$$\small \frac{d^2 i}{dt^2}+\frac{i}{LC}=0 $$

Solution is of the form \$\small i=Asin(\omega t+\phi)\$.

Substituting for \$\small i\$ in the ODE

$$\small -A\omega ^2 sin(\omega t+\phi)+\frac{Asin(\omega t+\phi)}{LC}=0$$

Hence \$\small \omega^2 =\frac{1}{LC}\$

Since the \$\small LC\$ circuit is has no resistance, the phase angle between current and voltage is known, hence the value of \$\small A\$ can be found from the initial conditions.

Answer 2

First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course). Besides that the answer of @Chu is excellent.

Well, we are trying to analyze the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_1=\text{I}_2+\text{I}_3\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag2 $$

Substitute \$(2)\$ into \$(1)\$, in order to get:

$$ \begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_2}{\text{R}_4} \end{cases}\tag3 $$

Now, we can solve for \$\text{V}_1\$ and \$\text{I}_3\$:

• $$\text{V}_1=\frac{\text{R}_2\left(\text{R}_3+\text{R}_4\right)\text{V}_\text{i}}{\text{R}_2\left(\text{R}_3+\text{R}_4\right)+\text{R}_1\left(\text{R}_2+\text{R}_3+\text{R}_4\right)}\tag4$$
• $$\text{I}_3=\frac{\text{R}_2\text{V}_\text{i}}{\text{R}_2\left(\text{R}_3+\text{R}_4\right)+\text{R}_1\left(\text{R}_2+\text{R}_3+\text{R}_4\right)}\tag5$$

Where I used the following Mathematica-code to find \$(4)\$ and \$(5)\$:

In[1]:=FullSimplify[
 Solve[{I1 == I2 + I3, I1 == (Vi - V1)/R1, I2 == V1/R2, 
   I3 == (V1 - V2)/R3, I3 == V2/R4}, {I1, I2, I3, V1, V2}]]

Out[1]={{I1 -> ((R2 + R3 + R4) Vi)/(R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  I2 -> ((R3 + R4) Vi)/(R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  I3 -> (R2 Vi)/(R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  V1 -> (R2 (R3 + R4) Vi)/(R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  V2 -> (R2 R4 Vi)/(R2 (R3 + R4) + R1 (R2 + R3 + R4))}}

---

Now, applying this to your circuit we need to use (from now on I use the lower case letters for the function in the 'complex' s-domain where I used Laplace transform):

• $$\text{R}_2=\frac{1}{\text{sC}}\tag6$$
• $$\text{R}_4=\text{sL}\tag7$$
• The input voltage is a stable DC voltage equal to \$\hat{\text{u}}\$, so: $$\text{v}_\text{i}\left(\text{s}\right)=\frac{\hat{\text{u}}}{\text{s}}\tag8$$

So, we get for \$(4)\$ and \$(5)\$:

• $$\text{v}_1\left(\text{s}\right)=\frac{\frac{1}{\text{sC}}\cdot\left(\text{R}_3+\text{sL}\right)\cdot\frac{\hat{\text{u}}}{\text{s}}}{\frac{1}{\text{sC}}\cdot\left(\text{R}_3+\text{sL}\right)+\text{R}_1\left(\frac{1}{\text{sC}}+\text{R}_3+\text{sL}\right)}\tag9$$
• $$\text{i}_3\left(\text{s}\right)=\frac{\frac{1}{\text{sC}}\cdot\frac{\hat{\text{u}}}{\text{s}}}{\frac{1}{\text{sC}}\cdot\left(\text{R}_3+\text{sL}\right)+\text{R}_1\left(\frac{1}{\text{sC}}+\text{R}_3+\text{sL}\right)}\tag{10}$$

Now, when we use inverse Laplace transform we can see:

• $$\lim_{t\to\infty}\text{V}_1\left(t\right)=\frac{\text{R}_3\hat{\text{u}}}{\text{R}_1+\text{R}_3}\tag{11}$$
• $$\lim_{t\to\infty}\text{I}_3\left(t\right)=\frac{\hat{\text{u}}}{\text{R}_1+\text{R}_3}\tag{12}$$

Where I used the following Mathematica-codes to find \$(11)\$ and \$(12)\$:

In[2]:=R2 = 1/(s*c);
R4 = s*L;
Vi = u/s; FullSimplify[
 Limit[InverseLaplaceTransform[(R2 (R3 + R4) Vi)/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), s, t], t -> Infinity, 
  Assumptions -> c > 0 && L > 0 && u > 0 && R1 > 0 && R3 > 0]]

Out[2]=ConditionalExpression[(R3 u)/(R1 + R3), 
 1/Sqrt[-4 c L R1 (R1 + R3) + (L + c R1 R3)^2] \[Element] Reals && 
  Sqrt[-4 c L R1 (R1 + R3) + (L + c R1 R3)^2] > 0 && 
  Sqrt[-4 c L R1 (R1 + R3) + (L + c R1 R3)^2] < L + c R1 R3 && 
  L + c R1 R3 + Sqrt[-4 c L R1 (R1 + R3) + (L + c R1 R3)^2] > 0]

In[3]:=R2 = 1/(s*c);
R4 = s*L;
Vi = u/s; FullSimplify[
 Limit[InverseLaplaceTransform[(R2 Vi)/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), s, t], t -> Infinity, 
  Assumptions -> c > 0 && L > 0 && u > 0 && R1 > 0 && R3 > 0]]

Out[3]=ConditionalExpression[u/(R1 + R3), 
 1/Sqrt[-4 c L R1 (R1 + R3) + (L + c R1 R3)^2] \[Element] Reals && 
  Sqrt[-4 c L R1 (R1 + R3) + (L + c R1 R3)^2] > 0 && 
  Sqrt[-4 c L R1 (R1 + R3) + (L + c R1 R3)^2] < L + c R1 R3 && 
  L + c R1 R3 + Sqrt[-4 c L R1 (R1 + R3) + (L + c R1 R3)^2] > 0]

Now, when the switch closes we can use Faraday's law to write:

$$\text{V}_\text{C}\left(t\right)=-\text{V}_\text{L}\left(t\right)\tag{13}$$

Now, we know the voltage current relations:

• Capacitor: $$\text{I}_\text{C}\left(t\right)=\text{V}_\text{C}'\left(t\right)\cdot\text{C}\tag{14}$$
• Inductor: $$\text{V}_\text{L}\left(t\right)=\text{I}_\text{L}'\left(t\right)\cdot\text{L}\tag{15}$$

And the circuit is in series so \$\text{I}\left(t\right):=\text{I}_\text{C}\left(t\right)=\text{I}_\text{L}\left(t\right)\$. And we know the initial conditons:

• $$\text{I}\left(0\right)=\frac{\hat{\text{u}}}{\text{R}_1+\text{R}_3}\tag{16}$$
• $$\text{V}_\text{C}\left(0\right)=\frac{\text{R}_3\hat{\text{u}}}{\text{R}_1+\text{R}_3}=-\text{I}'\left(0\right)\cdot\text{L}\tag{17}$$

So, we can solve for the current:

$$\text{I}\left(t\right)=\frac{\hat{\text{u}}}{\sqrt{\text{L}}}\cdot\frac{\sqrt{\text{L}}\cos\left(\frac{t}{\sqrt{\text{CL}}}\right)-\text{R}_3\sqrt{\text{C}}\sin\left(\frac{t}{\sqrt{\text{CL}}}\right)}{\text{R}_1+\text{R}_3}\tag{18}$$

Where I used the following Mathematica-code to find \$(18)\$:

In[4]:=DSolve[{x''[t]*L + (1/c)*x[t] == 0, 
  x[0] == u/(R1 + R3), (R3 u)/(R1 + R3) == -x'[0]*L}, x[t], t]

Out[4]={{x[t] -> -((
    u (-Sqrt[L] Cos[t/(Sqrt[c] Sqrt[L])] + 
       Sqrt[c] R3 Sin[t/(Sqrt[c] Sqrt[L])]))/(Sqrt[L] (R1 + R3)))}}