Finding the waveform of current in a circuit

Question

How do you find the waveform of current i1(t) in the following?

Do I need to find the capacitive and inductive reactance then find the current i1(t)?

Answer 1

If you actually mean the “waveform” of the current, then it will be a sinusoidal wave of the same frequency as the voltage input except scaled and shifted by some amount. If you look at a phasor diagram of current and voltage of any circuit, you will notice that the current always has the same waveform as the voltage. If you’re looking for the actual value(ie amplitude and phase shift) of the current, then yes you find the reactance and solve for i1.

Answer 2

Well, first of all, we can use a few standard things:

• Series capacitors can be added like resistors in parallel;
• Parallel capacitors can be added;
• Series coils can be added;
• Parallel coils be can added like resistors in parallel;
• $$\text{j}^2=-1\tag1$$
• $$\underline{\text{Z}}_{\space\text{C}}=\frac{1}{\text{j}\omega\text{C}}\tag2$$
• $$\underline{\text{Z}}_{\space\text{L}}=\text{j}\omega\text{L}\tag3$$
• $$\omega=2\pi\text{f}\tag4$$

Your circuit can be redrawn as follows:

^{simulate this circuit – Schematic created using CircuitLab}

Now, in your circuit we have:

$$\underline{\text{Z}}_{\space\text{in}}=\underline{\text{Z}}_{\space\text{C}}||\underline{\text{Z}}_{\space\text{L}}=\frac{\frac{1}{\text{j}\omega\text{C}}\cdot\text{j}\omega\text{L}}{\frac{1}{\text{j}\omega\text{C}}+\text{j}\omega\text{L}}=\frac{\omega\text{L}}{1-\omega^2\text{CL}}\cdot\text{j}\tag5$$

And we know that the complex input voltage (that is provided by the source) is given by:

$$\underline{\text{V}}_{\space\text{in}}=\hat{\text{v}}_{\space\text{in}}\exp\left(\varphi\cdot\text{j}\right)\tag6$$

Now, we can calculate the complex input current (that is provided by the source):

$$\underline{\text{I}}_{\space\text{in}}=\frac{\underline{\text{V}}_{\space\text{in}}}{\underline{\text{Z}}_{\space\text{in}}}=\frac{\hat{\text{v}}_{\space\text{in}}\exp\left(\varphi\cdot\text{j}\right)}{\left(\frac{\omega\text{L}}{1-\omega^2\text{CL}}\cdot\text{j}\right)}=\hat{\text{v}}_{\space\text{in}}\exp\left(\varphi\cdot\text{j}\right)\cdot\frac{\omega^2\text{CL}-1}{\omega\text{L}}\cdot\text{j}\tag7$$

Now, we know that the time representation of the input current is given by:

$$\text{I}_{\space\text{in}}\left(t\right)=\left|\underline{\text{I}}_{\space\text{in}}\right|\cos\left(\omega t+\arg\left(\underline{\text{I}}_{\space\text{in}}\right)\right)\tag8$$

Where:

• $$\left|\underline{\text{I}}_{\space\text{in}}\right|=\left|\hat{\text{v}}_{\space\text{in}}\exp\left(\varphi\cdot\text{j}\right)\cdot\frac{\omega^2\text{CL}-1}{\omega\text{L}}\cdot\text{j}\right|=\hat{\text{v}}_{\space\text{in}}\cdot\frac{\left|\omega^2\text{CL}-1\right|}{\omega\text{L}}\tag9$$
• $$\arg\left(\underline{\text{I}}_{\space\text{in}}\right)=\arg\left(\hat{\text{v}}_{\space\text{in}}\exp\left(\varphi\cdot\text{j}\right)\cdot\frac{\omega^2\text{CL}-1}{\omega\text{L}}\cdot\text{j}\right)$$ $$\arg\left(\hat{\text{v}}_{\space\text{in}}\right)+\arg\left(\exp\left(\varphi\cdot\text{j}\right)\right)+\arg\left(\left(\omega^2\text{CL}-1\right)\text{j}\right)-\arg\left(\omega\text{L}\right)=$$ $$0+\varphi+\arg\left(\left(\omega^2\text{CL}-1\right)\text{j}\right)-0=\varphi+\arg\left(\left(\omega^2\text{CL}-1\right)\text{j}\right)=$$ $$ \varphi+\begin{cases} 0,\space\space\space\space\space\space\space\space\space\space\space\space\omega^2\text{CL}-1=0\\ \\ \frac{\pi}{2},\space\space\space\space\space\space\space\space\space\space\space\omega^2\text{CL}-1>0\\ \\ \frac{3\pi}{2},\space\space\space\space\space\space\space\space\space\space\omega^2\text{CL}-1<0 \end{cases}\tag{10}$$

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Using your values:

• $$\text{C}=0.2\mu\text{F}||\left(3\space\text{nF}+0.047\space\mu\text{F}\right)=\frac{1}{25000000}\tag{11}$$
• $$\text{L}=75\mu\text{H}||\left(40\space\mu\text{H}+0.01\space\text{mH}\right)=\frac{3}{100000}\tag{12}$$
• $$\omega=10^6\tag{13}$$
• $$\hat{\text{v}}_{\space\text{in}}=6\tag{14}$$
• $$\varphi=-\frac{\pi}{2}\tag{15}$$

So:

• $$\left|\underline{\text{I}}_{\space\text{in}}\right|=6\cdot\frac{\left(10^6\right)^2\cdot\frac{1}{25000000}\cdot\frac{3}{100000}-1}{10^6\cdot\frac{3}{100000}}=\frac{1}{25}\tag{16}$$
• $$\arg\left(\underline{\text{I}}_{\space\text{in}}\right)=-\frac{\pi}{2}+\frac{\pi}{2}=0\tag{17}$$

We get for the input current:

$$\text{I}_{\space\text{in}}\left(t\right)=\frac{1}{25}\cdot\cos\left(10^6t\right)\tag{18}$$