LTspice value of peak current and theoretical value in a circuit question

Question

For the above question, after solving the circuit in frequency domain, I got the value of $$i_0(t) = 1.14\cos(3000t-55.3) mA $$ approximately (The calculations in the middle of steps was not approximated.)

Now, when I stimulate this in LTspice I get the following: The peak is approximately 1.056 mA of the current. There is a slight difference between the theoretical value and the value LTspice displays.

This did not happen only in this question, but many questions, where there always is a difference of about the second digit. (For instance, 840 V vs 865 V)

The circuit that was simulated:

So my question is, what is causing the difference and how is it possible to fix it?

Answer 1

Well, we are trying the 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:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_1+\text{I}_5&=\text{I}_2+\text{I}_3\\ \\ \text{I}_1&=\text{I}_2+\text{I}_4\\ \\ \text{I}_3&=\text{I}_4+\text{I}_5 \end{alignat*} \end{cases}\tag1 $$

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

$$ \begin{cases} \begin{alignat*}{1} \text{I}_1&=\frac{\displaystyle\text{V}_\text{i}-\text{V}_1}{\displaystyle\text{R}_1}\\ \\ \text{I}_2&=\frac{\displaystyle\text{V}_1-0}{\displaystyle\text{R}_2}\\ \\ \text{I}_3&=\frac{\displaystyle\text{V}_1-\text{V}_2}{\displaystyle\text{R}_3}\\ \\ \text{I}_3&=\frac{\displaystyle\text{V}_2-0}{\displaystyle\text{R}_4}\\ \\ \text{I}_5&=\frac{\displaystyle\text{n}\cdot\text{V}_2-\text{V}_1}{\displaystyle\text{R}_5} \end{alignat*} \end{cases}\tag2 $$

Now, we can subsitute \$\displaystyle\left(2\right)\$ into \$\displaystyle\left(1\right)\$ in order to rewrite \$\displaystyle\left(1\right)\$:

$$ \begin{cases} \begin{alignat*}{1} \frac{\displaystyle\text{V}_\text{i}-\text{V}_1}{\displaystyle\text{R}_1}+\frac{\displaystyle\text{n}\cdot\text{V}_2-\text{V}_1}{\displaystyle\text{R}_5}&=\frac{\displaystyle\text{V}_1-0}{\displaystyle\text{R}_2}+\frac{\displaystyle\text{V}_1-\text{V}_2}{\displaystyle\text{R}_3}\\ \\ \frac{\displaystyle\text{V}_\text{i}-\text{V}_1}{\displaystyle\text{R}_1}+\frac{\displaystyle\text{n}\cdot\text{V}_2-\text{V}_1}{\displaystyle\text{R}_5}&=\frac{\displaystyle\text{V}_1-0}{\displaystyle\text{R}_2}+\frac{\displaystyle\text{V}_2-0}{\displaystyle\text{R}_4}\\ \\ \frac{\displaystyle\text{V}_\text{i}-\text{V}_1}{\displaystyle\text{R}_1}&=\frac{\displaystyle\text{V}_1-0}{\displaystyle\text{R}_2}+\text{I}_4\\ \\ \frac{\displaystyle\text{V}_1-\text{V}_2}{\displaystyle\text{R}_3}&=\text{I}_4+\frac{\displaystyle\text{n}\cdot\text{V}_2-\text{V}_1}{\displaystyle\text{R}_5}\\ \\ \frac{\displaystyle\text{V}_2-0}{\displaystyle\text{R}_4}&=\text{I}_4+\frac{\displaystyle\text{n}\cdot\text{V}_2-\text{V}_1}{\displaystyle\text{R}_5} \end{alignat*} \end{cases}\tag3 $$

Now, you can use Laplace transform to write:

$$\text{R}_2=\frac{\displaystyle1}{\displaystyle\text{sC}},\space\text{R}_3=\text{sL}\tag4$$

And:

$$\text{v}_\text{i}\left(\text{s}\right)=\mathscr{L}_t\left[\hat{\text{u}}_\text{i}\cos\left(\omega t\right)\right]_{\left(\text{s}\right)}=\frac{\displaystyle\hat{\text{u}}_\text{i}\text{s}}{\displaystyle\text{s}^2+\omega^2}\tag5$$

Now, for your case you need to solve for \$\displaystyle-\text{I}_5\$ in order to find your \$\displaystyle i_o\$.