Steady State Analysis of 'Complex' Circuit

Electricity and Magnetism Level 1

Consider the arrangement above.

V S = V o sin ⁡ ( ω t ) V_S = V_o \sin(\omega t)

Let the circuit impedance be Z ( ω ) Z(\omega) . Enter your answer as:

lim ⁡ ω → ∞ ∣ Z ( ω ) ∣ \lim_{\omega \to \infty} \lvert Z(\omega)\rvert

Note: Here ∣ Z ( ω ) ∣ \lvert Z(\omega)\rvert represents the magnitude of the impedance.

This problem is inspired by a comment from Talulah Riley.


The answer is 10.

Karan Chatrath by Karan Chatrath , 27, Netherlands

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1 solution

Steven Chase
Oct 13, 2020

As the source angular frequency tends toward infinity, the capacitor behaves like a short circuit. This is because its impedance has the following form:

Z C = 1 j ω C Z_C = \frac{1}{j \omega C}

When the capacitor is a short, only the 10 Ω 10 \Omega resistor is effectively still in the circuit

2 Helpful 2 Interesting 2 Brilliant 0 Confused

@Steven Chase at t = ∞ t=\infty capacitor will disconnect. So the net resistance will be 40 Ω 40 \Omega .
Where I am wrong. Please correct me. Thanks in advance.

Talulah Riley - 8 months ago

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At infinite angular frequency, the capacitor behaves as a short circuit.

Steven Chase - 8 months ago

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