Complex Pi Circuit Analysis

Electricity and Magnetism Level 3

A π \pi -circuit consists of two shunt capacitive impedances of − j Ω -j \Omega and one series inductive impedance of j Ω j \Omega .

An ideal AC voltage source feeds the circuit with V ⃗ \vec{V} volts and I ⃗ \vec{I} amps.

What is the value of the load impedance Z ⃗ ? \vec{Z}?

Note: Regardless of which expression you pick, assume that Z ⃗ \vec{Z} has units of ohms.

Clarification: j = − 1 . j = \sqrt{-1}.

− j I ⃗ V ⃗ {\frac{-j \vec{I}}{\vec{V}}} j V ⃗ I ⃗ {\frac{j \vec{V}}{\vec{I}}} V ⃗ I ⃗ {\frac{\vec{V}}{\vec{I}}} I ⃗ V ⃗ {\frac{\vec{I}}{\vec{V}}}

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Steven Chase by Steven Chase , 37, USA

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1 solution

Steven Chase
Feb 3, 2017

Note: Given the final result, you may be wondering how something that looks like admittance can have units of ohms. This is because of the transformation 1 j = − j \frac{1}{j} = -j . The " j j " impedances in the problem have units of ohms, and when they are transformed, their transformations have units of inverse ohms, or siemens. For the sake of keeping things uncluttered, I have neglected to catalog all of that meticulously. However, numerical calculations readily confirm that the end result is correct.

The current through the first capacitor is: I C 1 ⃗ = V ⃗ − j = j V ⃗ . \large{\vec{I_{C1}} = \frac{\vec{V}}{-j} = j\vec{V}}.

The current through the inductor is therefore: I L ⃗ = I ⃗ − j V ⃗ . \large{\vec{I_L} = \vec{I} - j\vec{V}}.

The voltage at the load is: V L o a d ⃗ = V ⃗ − j ( I ⃗ − j V ⃗ ) = V ⃗ − j I ⃗ − V ⃗ = − j I ⃗ . \large{\vec{V_{Load}} = \vec{V} - j(\vec{I} - j\vec{V}) = \vec{V} -j\vec{I} - \vec{V} = -j\vec{I}}.

The current through the second capacitor is:
I C 2 ⃗ = V L o a d ⃗ − j = − j I ⃗ − j = I ⃗ . \large{\vec{I_{C2}} = \frac{\vec{V_{Load}}}{-j} = \frac{-j\vec{I}}{-j} = \vec{I}}.

The current through the load is therefore:
I L o a d ⃗ = I L ⃗ − I C 2 ⃗ = I ⃗ − j V ⃗ − I ⃗ = − j V ⃗ . \large{\vec{I_{Load}} = \vec{I_L} - \vec{I_{C2}} = \vec{I} - j\vec{V} - \vec{I} = -j\vec{V}}.

The load impedance is therefore:
Z ⃗ = V L o a d ⃗ I L o a d ⃗ = − j I ⃗ − j V ⃗ = I ⃗ V ⃗ . \large{\vec{Z} = \frac{\vec{V_{Load}}}{\vec{I_{Load}}} = \frac{-j\vec{I}}{-j\vec{V}} = \boxed{\frac{\vec{I}}{\vec{V}}}}.

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