Impedance of a series two-component circuit

This is an AC circuit operating at a frequency of 1 MHz 2 π \frac{1\text{MHz}}{2\pi} . The inductors are 1 mH and the capacitors are 1 nF.

Simplifying assumptions: the components are pure, ideal components, there is no mutual inductance between the inductors and the wiring is negligible.

The device impedance formulae are simple and the frequency was chosen to give integer impedance magnitudes, in the sense that a unit circle in the complex plane has a magnitude of 1 1 around the entire circle. The standard unit for inductors is a Henry and for capacitors is a Farad. Convert the units used above to those units before computing the device impedances.

Here are the necessary formulas.


The answer is 0.

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2 solutions

1000 i + 1000 i = = 0 1000i+-1000i==0 Yes, it is really that simple.

Inductor and capacitor reactances are equal at given frequency(LC resonance), so the total impedance is 0.

Your answer is correct. Your statement about the reactances is wrong. They are equal in magnitude and opposite sign. 2+2=4 not 0. 2+-2=0. Even most textbooks get this wrong. I tried for years to get the American Radio Relay League Handbook corrected on this topic.

A Former Brilliant Member - 2 years, 4 months ago

Thank you for correcting me. You are totally right about textbooks, I learned that the reactance of a capacitor is always a positive real number, but its impedance is an imaginary number with a minus sign.

Novak Radivojević - 2 years, 4 months ago

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