< Practical Electronics

Parallel RC

Circuit Impedance

Circuit Response

Parallel RL

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Circuit Impedance

Circuit Response

Parallel LC

Circuit Impedance

Circuit response

Parallel RLC

Circuit Impedance

Circuit response

Natural Respond

Forced Respond

Second ordered equation that has two roots

ω = -α ±

Where

The current of the network is given by

A eω1 t + B eω2 t

From above

When , there is only one real root
ω = -α
When , there are two real roots
ω = -α ±
When , there are two complex roots
ω = -α ± j

Resonance Response

At resonance, the impedance of the frequency dependent components cancel out . Therefore the net voltage of the circui is zero

and

At Resonance Frequency

.
. Current is at its maximum value

Further analyse the circuit

At ω = 0, Capacitor Opened circuit . Therefore, I = 0 .
At ω = 00, Inductor Opened circuit . Therefore, I = 0 .


With the values of Current at three ω = 0 , , 00 we have the plot of I versus ω . From the plot If current is reduced to halved of the value of peak current , this current value is stable over a Frequency Band ω1 - ω2 where ω1 = ωo - Δω, ω2 = ωo + Δω


  • In RLC series, it is possible to have a band of frequencies where current is stable, ie. current does not change with frequency . For a wide band of frequencies respond, current must be reduced from it's peak value . The more current is reduced, the wider the bandwidth . Therefore, this network can be used as Tuned Selected Band Pass Filter . If tune either L or C to the resonance frequency . Current is at its maximum value . Then, adjust the value of R to have a value less than the peak current by increasing R to have a desired frequency band .


  • If R is increased from R to 2R then the current now is which is stable over a band of frequency
ω1 - ω2 where
ω1 = ωo - Δω
ω2 = ωo + Δω

For value of I < . The circuit respond to Wide Band of frequencies . For value of < I > . The circuit respond to Narrow Band of frequencies

Summary

CircuitSymbolSeriesParallel
RC
A parallel RC Circuit
ImpedanceZ
Frequency



VoltageV
CurrentI
Phase AngleTan θ = 1/2πf RC
f = 1/2π Tan CR
t = 2π Tan CR
Tan θ = 1/2πf RC
f = 1/2π Tan CR
t = 2π Tan CR
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