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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+...`

Continues below

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):

`=((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"`

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^@`.

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