Simple Circuit with complex numbers

Simple Circuit

Sample problem #1

Find current I.

We start solving this circuit by converting elements into impedances.

(W is the angular frequency which is given from the voltage source.)

W= 30

For the 4 H inductor we will use the formula z=jWL:

Zl= j(30)(200mH)

                                                                    Zl= j6 ohm

And for the 3m F capacitor, we will use z= 1/jWc.

                                                                     Zc=1/jWc

Zc=1/j(30)(3mF)

                                                                     Zc= 1/j0.09

Zc=-j11.11 ohm

Substitute the solved impedance on the circuit.

We could see that the impedances could be combined. We will use the product over sum rule and simple addition to combine the three elements.

Zl and Zc are parallel:

Z1= (j6) (-j11.11)/j6-j11.11

                                                            Z1= j13.045 ohm

Then, Z1 and ZR are in series so we just add them.

                                                                   Z= j13.045+ 1k

Z= 1000 +j13.045

The simplified circuit would be like this:

To get the unknown current we could simply apply Ohms law:

I= V/R

I= 3.064 -j2.571/ 1000 +j13.045

                                                                   I= 3.03 –j2.610 mA

Converting it to polar form would result to:

I= 3.999 ∟-40.75 mA