# Practical Electronics/Parallel RC

## Parallel RC

### Circuit Impedance

$\frac {1}{Z} = \frac{1}{Z_R} + \frac{1}{Z_C}$
$\frac{1}{Z} = \frac{1}{R} + j\omega C = \frac {j\omega CR + 1}{R}$
$Z = R \frac {1}{j\omega CR + 1}$

### Circuit Response

$I = I_R + I_C$
$I = \frac {V}{R} + C \frac{dV}{dt}$
$V = (IR - RC \frac {dV}{dt})$

## Parallel RL

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### Circuit Impedance

$\frac{1}{Z} = \frac{1}{Z_R} + \frac{1}{Z_L}$
$\frac{1}{Z} = \frac{1}{R} + \frac{1}{j\omega L} = \frac{R + j\omega L}{j\omega RL}$
$Z = \frac{j\omega RL}{R + j\omega L} = j\omega L\frac {1} {1 + j\omega \frac{L}{R}}$

### Circuit Response

$I = I_R + I_L$
$I = \frac{V}{R} + \frac{1}{L} \int V dt$
$V = IR - \frac{R}{L} \int V dt$

## Parallel LC

### Circuit Impedance

$\frac{1}{Z} = \frac{1}{Z_L} + \frac{1}{Z_C}$
$\frac{1}{Z} = \frac{1}{j\omega L} + j\omega C = \frac{(j\omega)^2 LC + 1}{j\omega L}$
$Z = \frac{(j\omega)^2 LC + 1}{j\omega L}$

### Circuit response

$I = I_L + I_C$
$I = \frac{1}{L} \int V dt + C \frac{dV}{dt}$

## Parallel RLC

### Circuit Impedance

$\frac{1}{Z} = \frac{1}{Z_R} + \frac{1}{Z_L} + \frac{1}{Z_C}$
$\frac{1}{Z} = \frac{1}{R} + \frac{1}{j\omega L} + j\omega C$
$\frac{1}{Z} = \frac{(j\omega)^2 LC + j\omega L + R }{j\omega RL}$
$\frac{1}{Z} = \frac{(j\omega)^2 \frac{LC}{R} + j\omega \frac{L}{R} + 1 }{j\omega L}$

### Circuit response

$I = I_R + I_L + I_C$
$I = \frac{V}{R} + \frac{1}{L} \int V dt + C \frac{dV}{dt}$
$I = \frac{V}{R} + \frac{1}{L} \int V dt + C \frac{dV}{dt}$
$V = IR - \frac{R}{L} \int V dt - CR \frac{dV}{dt}$

#### Natural Respond

$0 = \frac{V}{R} + \frac{1}{L} \int V dt + C \frac{dV}{dt}$

#### Forced Respond

$I_t = IR + L \frac{dI}{dt} + \frac{1}{C} \int I dt$

Second ordered equation that has two roots

ω = -α ± $\sqrt {\alpha^2 - \beta^2}$

Where

$\alpha = \frac{R}{2L}$
$\beta = \frac{1}{\sqrt{LC}}$

The current of the network is given by

A eω1 t + B eω2 t

From above

When ${\alpha^2 = \beta^2}$, there is only one real root
ω = -α
When ${\alpha^2 > \beta^2}$, there are two real roots
ω = -α ± $\sqrt {\alpha^2 - \beta^2}$
When ${\alpha^2 < \beta^2}$, there are two complex roots
ω = -α ± j$\sqrt {\beta^2 - \alpha^2 }$

#### Resonance Response

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

$Z_L - Z_C = 0$ and $V_L + V_C = 0$

$\omega L = \frac{1}{\omega C}$
$\omega = \sqrt {\frac{1}{LC}}$
$Z = Z_R + (Z_L - Z_C) = Z_R = R$
$I = \frac{V}{R}$

At Resonance Frequency

$\omega = \sqrt {\frac{1}{LC}}$ .
$I = \frac{V}{R}$ . 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 , $\sqrt {\frac{1}{LC}}$ , 00 we have the plot of I versus ω . From the plot If current is reduced to halved of the value of peak current $I = \frac{V}{2R}$ , 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 $\omega = \sqrt {\frac{1}{LC}}$ . Current is at its maximum value $I = \frac{V}{R}$ . Then, adjust the value of R to have a value less than the peak current $I = \frac{V}{R}$ by increasing R to have a desired frequency band .

• If R is increased from R to 2R then the current now is $I = \frac{V}{2R}$ which is stable over a band of frequency
ω1 - ω2 where
ω1 = ωo - Δω
ω2 = ωo + Δω

For value of I < $I = \frac{V}{2R}$ . The circuit respond to Wide Band of frequencies . For value of $I = \frac{V}{R}$ < I > $I = \frac{V}{2R}$ . The circuit respond to Narrow Band of frequencies

## Summary

Circuit Symbol Series Parallel
RC
A parallel RC Circuit
Impedance Z $Z_t = R + \frac {1}{\omega C} = \frac {\omega CR + 1}{\omega C}$ $\frac{1}{Z_t} = \frac{1}{Z_R} + \frac{1}{Z_C} = \frac{1}{R} + \omega C = \frac {R}{\omega CR + 1}$
Frequency $\omega_o = 2 f_o$ $Z_R = Z_C$
$R = \frac{1}{\omega C}$
$\omega = \frac{1}{CR}$
$\frac{1}{R} = \frac{1}{\omega C}$
$\frac{1}{R} = \omega C$
$\omega = \frac{1}{CR}$
Voltage V $V = IR + \frac {1}{C} \int I dt$ $I = \frac {V}{R} + C\frac{dV}{dt}$
Current I $\int I dt = C (V - IR)$ $\frac {dV}{dt} = \frac{1}{C}(I - \frac{V}{R})$
Phase Angle Tan 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