Impedances in Parallel

Given two impedances Z1 and Z2 connected in parallel, what is their impedance expressed in real and imaginary parts?

For impedances in parallel, the formula is:

\frac{1}{Z} = \frac{1}{Z_{1}} + \frac{1}{Z_{2}}

Expressing each impedance Z1 and Z2 individually as:

MathML formula	Formula Image	Straight text formula
$Z_{1} = a + b j$		Z₁ = a + bj
$Z_{2} = c + d j$		Z₂ = c + dj

Here is the online calculator tool:

(Note: Polar notation using angle symbol ∠Ð to be added later)

(Remember that the imaginary part is positive for inductive reactance, and negative for capacitive reactance)

Formula Derivation

We can now derive an expression for the combined impedance Z:

\frac{1}{Z} = \frac{Z_{1} + Z_{2}}{Z_{1} Z_{2}}

\Rightarrow Z = \frac{Z_{1} Z_{2}}{Z_{1} + Z_{2}}

\Rightarrow Z = \frac{(a + b j) (c + d j)}{(a + b j) + (c + d j)}

Re-arranging the denominator gives:

\Rightarrow Z = \frac{(a + b j) (c + d j)}{(a + c) + (b + d) j}

Now, multiplying above and below by:

[(a + c) - (b + d) j]

Gives Us:

Z = \frac{[(a + b j) (c + d j)] [(a + c) - (b + d) j]}{[(a + c) + (b + d) j] [(a + c) - (b + d) j]}

Now, recognising that:

j^{2} = -1

We can expand to give:

Z = \frac{[a c + a d j + b c j - b d] [a + c - b j - d j]}{{(a + c)}^{2} + {(b + d)}^{2}}

\Rightarrow Z = \frac{a^{2} c + a c^{2} - a b c j - a c d j + a^{2} d j + a c d j - a b d j^{2} - a d^{2} j^{2} + a b c j + b c^{2} j - b^{2} c j^{2} - b c d j^{2} - a b d - b c d + b^{2} d j + b d^{2} j}{{(a + c)}^{2} + {(b + d)}^{2}}

\Rightarrow Z = \frac{a^{2} c + a c^{2} - a b c j - a c d j + a^{2} d j + a c d j + a b d + a d^{2} + a b c j + b c^{2} j + b^{2} c + b c d - a b d - b c d + b^{2} d j + b d^{2} j}{{(a + c)}^{2} + {(b + d)}^{2}}

\Rightarrow Z = \frac{a^{2} c + a c^{2} + a b d + a d^{2} + b^{2} c + b c d - a b d - b c d - a b c j - a c d j + a^{2} d j + a c d j + a b c j + b c^{2} j + b^{2} d j + b d^{2} j}{{(a + c)}^{2} + {(b + d)}^{2}}

\Rightarrow Z = \frac{a^{2} c + a c^{2} + a d^{2} + b^{2} c + (a^{2} d + b c^{2} + b^{2} d + b d^{2}) j}{{(a + c)}^{2} + {(b + d)}^{2}}

So, the solution is:

MathML formula
$\Rightarrow Z = \frac{a^{2} c + a c^{2} + a d^{2} + b^{2} c}{{(a + c)}^{2} + {(b + d)}^{2}} + \frac{(a^{2} d + b c^{2} + b^{2} d + b d^{2})}{{(a + c)}^{2} + {(b + d)}^{2}} j$
Formula Image

Straight text formula
Z = [ [a²c + ac² + ad² + b²c] / [(a + c)² + (b + d)²] ] + [ [a²d + bc² + b²d + bd²] / [(a + c)² + (b + d)²] ] j