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

Tips

-Solving complex number is much more easier with the use of  scientific calculators especially those with      complex functions.

-Don't get confused of the use of complex numbers in circuit analysis 2. Just think that we are solving the same problems and  using the same methods in circuit analysis 1 but the  given real numbers in the problems are changed to  complex numbers.

- .........