Skip to main content

12. Parallel AC Circuits

Recall Ohm's law for pure resistances:

`V = IR`

In the case of AC circuits, we represent the impedance (effective resistance) as a complex number, Z. The units are ohms (`Ω`).

In this case, Ohm's Law becomes:

V = IZ.

Recall also, if we have several resistors (R1, R2, R3, R4, …) connected in parallel, then the total resistance RT, is given by:

`1/(R_T)=1/R_1+1/R_2+1/R_3+...`

In the case of AC circuits, this becomes:

`1/(Z_T)=1/Z_1+1/Z_2+1/Z_3+...`

Simple case:

If we have 2 impedances Z1 and Z2, connected in parallel, then the total resistance ZT, is given by

`1/(Z_T)=1/Z_1+1/Z_2`

We can write this as:

`1/(Z_T)=(Z_2+Z_1)/(Z_1Z_2)`

Finding the reciprocal of both sides gives us:

`Z_T=(Z_1Z_2)/(Z_1+Z_2)`

Example 1

Find the combined impedance of the following circuit:

Circuit diagram

Answer

Call the impedance given by the top part of the circuit Z1 and the impedance given by the bottom part Z2.

We see that Z1 = 70 + 60j Ω and Z2 = 40 − 25j Ω

So

`Z_T=(Z_1Z_2)/(Z_1+Z_2)`

`=((70+60j)(40-25j))/((70+60j)+(40-25j))`

`=((70+60j)(40-25j))/(110+35j)`

(Adding complex numbers should be done in rectangular form.)

Now, we convert everything to polar form and then multiply and divide as follows:

`Z_T =((70+60j)(40-25j))/(110+35j)`

`=((92.20/_40.60^text(o))(47.17/_-32.01^text(o)))/(115.4/_17.65^text(o))`

(We do the product on the top first.)

`=((92.20xx47.17)/_(40.60^text(o)-32.01^text(o)))/(115.4/_17.65^text(o))`

`=(4349.074/_8.59^text(o))/(115.4/_17.65^text(o))`

(Now we do the division.)

`=(4349.074)/115.4/_(8.59^text(o)-17.65^text(o))`

`=37.69/_-9.06^text(o)`

(We convert back to rectangular form.)

`=37.22-5.93j`

(When multiplying complex numbers in polar form, we multiply the r terms (the numbers out the front) and add the angles. When dividing complex numbers in polar form, we divide the r terms and subtract the angles. See the Products and Quotients section for more information.)

So we conclude that the combined impedance is

`Z_T = 37-5.9j\ Omega`

Example 2

Given that Z1= 200 − 40j Ω and Z2= 60 + 130j Ω,

Circuit diagram

find

a) the total impedance

b) the phase angle

c) the total line current

Answer

a) `Z_T =frac{Z_1Z_2}{Z_1 + Z_2}`

`=frac{(200-40j)(60+130j)}{(200-40j)+(60+130j)} `

`=frac{(200-40j)(60+130j)}{260+90j}`

`=frac{(204.0angle-11.31^@)(143.2angle65.22^@)}{(275.1angle19.09^@)}`

`=frac{204.0times143.2}{275.1}angle(-11.31^@+` `65.22^@-` `{:19.09^@)`

`=106.2angle34.82^@`

`=87.18+60.64j`

So we conclude that the total impedance is

`Z_T = 87.2+60.6j\ Omega`


b) We see from the second last line of our last answer that the phase angle is `~~35^@`.


c) Total line current:

We use

  • V = IZ,
  • the fact that the impedance is `106.2 ∠ 34.82^@` and
  • the fact that the voltage supplied is `12 V = 12 ∠ 0^@\ V.`

So

`I = V/Z`

`=frac{12angle0^@}{106.2angle34.82^@}`

`=0.113angle-34.82^@ "A"`

Easy to understand math videos:
MathTutorDVD.com

Example 3

A `100\ Ω` resistor, a `0.0200\ "H"` inductor and a `1.20\ mu"F"` capacitor are connected in parallel with a circuit made up of a `110\ Ω` resistor in series with a `2.40\ mu"F"` capacitor. A supply of `150\ "V"`, `60\ "Hz"` is connected to the circuit.

Calculate the total current taken from the supply and its phase angle.

Answer

Parallel RLC circuit diagram - application of complex numbers


For Z1 (the upper part of the circuit), we have:

XL = 2πfL = 2π (60)(0.0200) = 7.540 Ω

`X_C=1/(2pi(60)(1.20xx10^-6))`

`=2210.485\ Omega`

Z1 = R1 + j(XLXC)

= 100 + j(7.540 - 2210.485)

= 100 − 2202.9j

`= 2205.21 ∠ − 87.40^@\ Ω`

For Z2 (the lower part of the circuit), we have:

`X_C=1/(2pi(60)(2.40xx10^-6))`

`=1105.243\ Omega`

Z2 = R2 + j(XL - XC)

= 110 + j(−1105.243)

`= 1110.7 ∠ −84.32^@\ Ω`

So the total impedance, ZT, is given by:

`Z_T=(Z_1Z_2)/(Z_1+Z_2)`

`=(2449326.75/_-171.72^"o")/(210-3308.188j)`

`=(2449326.75/_-171.72^"o")/(3314.85/_-86.37^"o")`

`=738.9/_-85.35^"o"`

This last line in rectangular form is ZT = 59.9 − 736.5j Ω

Now:

`I_T=(V_T)/(Z_T)`

`=(150/_0^"o")/(738.9/_-85.35^"o")`

`=0.203/_85.35^"o"`

So the total current taken from the supply is `203\ "mA"` and the phase angle of the current is `~~85^@`.

top

Search IntMath, blog and Forum

Search IntMath

Online Algebra Solver

This algebra solver can solve a wide range of math problems.

Algebra Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand algebra lessons on DVD. See samples before you commit.

More info: Algebra videos

The IntMath Newsletter

Sign up for the free IntMath Newsletter. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!


See the Interactive Mathematics spam guarantee.