A large AC electric motor under load can be
considered as a parallel combination of resistance and inductance. If the
equivalent resistance is 20 Ω and the inductance is 238 mH, calculate the rms
current delivered by the 60 Hz, V_{rms} = 240 V AC power supply.

Solution:

- Concepts:

AC circuits - Reasoning:

We are analyzing a parallel RL circuit. - Details of the calculation:

I_{rms}= V_{rms}/|Z|.

Z = RZ_{L}/(R + Z_{L}) = (ω^{2}L^{2}R + iωLR^{2})/(R^{2}+ ω^{2}L^{2})

= ωLR(R^{2}+ ω^{2}L^{2})^{½}/(R^{2}+ ω^{2}L^{2})exp(iφ) = ωLR/(R^{2}+ ω^{2}L^{2})^{½}exp(iφ) = |Z|exp(iφ).

|Z| = 19.52 Ω.

I_{rms}= V_{rms}/|Z| = (240 V)/(19.52 Ω) = 12.3 A.

In an AC circuit, a filter circuit or filter removes unwanted frequency
components from an input. A passive filter consists of a circuit with at
least one reactive component. A filter with a single reactive component,
either a capacitor or an inductor, is called a first-order filter. Filters
are described according to their behavior. A low-pass filter will pass all
signals below a specific frequency, but will attenuate or block signals of a
higher frequency. A high-pass filter will do just the opposite. The
boundary line is somewhat arbitrary since there exists a "transition zone".
The cutoff frequency is defined as the frequency at which |V_{out}| = 2^{-½}|V_{in}|.

Inspect the filter shown below. Identify what type of filter this
circuit is, and calculate the size of resistor necessary to give it a cutoff
frequency of 3 kHz:

Solution:

- Concepts:

AC circuits - Reasoning:

We have a series RL circuit. - Details of the calculation:

Z = R + iωL.

We want V_{out}= V_{in}Z_{L}/Z = V_{in}exp(iφ)ωL/(R^{2}+ (ωL)^{2})^{½}= (1/√2)V_{in}exp(iφ).

We need R = ωL = 2π*3*10^{3}*3*10^{-1}Ω = 5.65 kΩ.

This is a high-pass filter.