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 |
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${Z}_{1}=a+bj$ | Z_{1} = a + bj | |

${Z}_{2}=c+dj$ | Z_{2} = c + dj |

Here is the online calculator tool:

Real Part | Imaginary Part | |
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Z_{1} | ||

Z_{2} | ||

Z_{T} |

(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:

$$\frac{1}{Z}=\frac{{Z}_{1}+{Z}_{2}}{{Z}_{1}{Z}_{2}}$$ $$\Rightarrow Z=\frac{{Z}_{1}{Z}_{2}}{{Z}_{1}+{Z}_{2}}$$ $$\Rightarrow Z=\frac{\left(a+bj\right)\left(c+dj\right)}{\left(a+bj\right)+\left(c+dj\right)}$$Re-arranging the denominator gives:

$$\Rightarrow Z=\frac{\left(a+bj\right)\left(c+dj\right)}{\left(a+c\right)+\left(b+d\right)j}$$Now, multiplying above and below by:

$$\left[\left(a+c\right)-\left(b+d\right)j\right]$$ Gives Us: $$Z=\frac{\left[\left(a+bj\right)\left(c+dj\right)\right]\left[\left(a+c\right)-\left(b+d\right)j\right]}{\left[\left(a+c\right)+\left(b+d\right)j\right]\left[\left(a+c\right)-\left(b+d\right)j\right]}$$Now, recognising that:

$${j}^{2}=-1$$We can expand to give:

$$Z=\frac{\left[ac+adj+bcj-bd\right]\left[a+c-bj-dj\right]}{{\left(a+c\right)}^{2}+{\left(b+d\right)}^{2}}$$ $$\Rightarrow Z=\frac{{a}^{2}c+a{c}^{2}-abcj-acdj+{a}^{2}dj+acdj-abd{j}^{2}-a{d}^{2}{j}^{2}+abcj+b{c}^{2}j-{b}^{2}c{j}^{2}-bcd{j}^{2}-abd-bcd+{b}^{2}dj+b{d}^{2}j}{{\left(a+c\right)}^{2}+{\left(b+d\right)}^{2}}$$ $$\Rightarrow Z=\frac{{a}^{2}c+a{c}^{2}-abcj-acdj+{a}^{2}dj+acdj+abd+a{d}^{2}+abcj+b{c}^{2}j+{b}^{2}c+bcd-abd-bcd+{b}^{2}dj+b{d}^{2}j}{{\left(a+c\right)}^{2}+{\left(b+d\right)}^{2}}$$ $$\Rightarrow Z=\frac{{a}^{2}c+a{c}^{2}+abd+a{d}^{2}+{b}^{2}c+bcd-abd-bcd-abcj-acdj+{a}^{2}dj+acdj+abcj+b{c}^{2}j+{b}^{2}dj+b{d}^{2}j}{{\left(a+c\right)}^{2}+{\left(b+d\right)}^{2}}$$ $$\Rightarrow Z=\frac{{a}^{2}c+a{c}^{2}+a{d}^{2}+{b}^{2}c+\left({a}^{2}d+b{c}^{2}+{b}^{2}d+b{d}^{2}\right)j}{{\left(a+c\right)}^{2}+{\left(b+d\right)}^{2}}$$So, the solution is:

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

Formula Image |

Straight text formula |

Z = [ [a^{2}c + ac^{2} + ad^{2} + b^{2}c] / [(a + c)^{2} + (b + d)^{2}] ] + [ [a^{2}d + bc^{2} + b^{2}d + bd^{2}] / [(a + c)^{2} + (b + d)^{2}] ] j |