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:

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

Using your values:

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

We get for the input current:

$$\text{I}_{\space\text{in}}\left(t\right)=\frac{1}{25}\cdot\cos\left(10^6t\right)\tag{16}$$

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:

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

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:

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

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:

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

Using your values:

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

We get for the input current:

$$\text{I}_{\space\text{in}}\left(t\right)=\frac{1}{25}\cdot\cos\left(10^6t\right)\tag{16}$$

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:

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

Using your values:

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

We get for the input current:

$$\text{I}_{\space\text{in}}\left(t\right)=\frac{1}{25}\cdot\cos\left(10^6t\right)\tag{16}$$