Drawing of the phasor diagram for a series RLC circuit energized by a sinusoidal voltage showing the relative position of current, component voltage and applied voltage for the following case

a)When X_L > Xc

b)When X_L < Xc

c)When X_L = Xc.

Drawing of the phasor diagram for a series RLC circuit energized by a sinusoidal voltage showing the relative position of current, component voltage and applied voltage for the following case. A) When X L Xc b) When X L RLC Circuit.
Oleh karena itu, hubungan tegangan lebih baik dijelaskan dengan menggunakan diagram fasor. Praktikum ini dilakukan untuk, memberikan pemahaman dan pengetahuan akan rangkaian RLC baik dari penerapannya maupun sifat-sifat rangkaian tersebut B. Tujuan Setelah melaksanakan percobaan, praktikan diharapkan dapat: 1.

Perhatikan diagram rangkaian RLC berikut ini! Kuat arus maksimum dari rangkaian ini adalah (1 pF = 10-6 F) A. Latihan 5 (UN 2013) Perhatikan gambar rangkaian listrik berikut! Jika tegangan maksimum sumber arus bolak¬balik = 200 V, maka besar kuat arus maksimum yang mengalir pada rangkaian adalah A. Hubungan impedansi dapat dinyatakan dengan diagram fasor yakni diagram vektor besaran dan sudut fase gelombang. Baca juga: Teknologi Digital Fisika. Rangkaian Arus Bolak-balik. Secara sederhana pada kehidupan sehari rangkaian bolak-balik dapat dihubungkan dengan resistor, induktor / selenoida, serta kapasitor. Diagram fasor rangkaian rlc paralel. Masih banyak lagi contoh dari penerapan rangkaian paralel pada lampu. Baca juga hukum ohm dan contoh soalnya. Pada rangkaian rlc paralel terjadi pembagian arus listrik dari sumber menjadi tiga yaitu menuju ke resistor induktor dan kapasitor.

RLC Circuit:

Consider a circuit in which R, L, and C are connected in series with each other across ac supply as shown in fig.

The ac supply is given by,

V = Vm sin wt

The circuit draws a current I. Due to that different voltage drops are,

1)Voltage drop across Resistance R is V_R = IR

2)Voltage drop across Inductance L is V_L = IX_L

3) Voltage drop across Capacitance C is Vc = IXc The characteristics of three drops are,

1.V_R is in phase with current I

2.V_L leads I by 90⁰

3.Vc lags I by 90⁰

According to krichoff’s laws

Steps to draw phasor diagram:

1.Take current I as reference

2.V_R is in phase with current I

3.V_L leads current by 90⁰

4.Vc lags current by 90⁰

5.obtain resultant of V_L and Vc. Both V_L and Vc are in phase opposition (180⁰ out of phase)

6.Add that with VRby law of parallelogram to getsupply voltage.

The phasor diagram depends on the condition of magnitude of V_L and Vc which ultimately depends on values of X_L and Xc.

Let us consider different cases:

Case(i): X_L > Xc

When X _L > Xc

Also V_L > Vc (or) IX_L > IXc

So, resultant of V_L and Vc will directed towards V_L i.e. leading current I. Hence I lags V i.e. current I will lags the resultant of V_L and Vc i.e. (V_L - Vc). The circuit is said to be inductive in nature.

From voltage triangle,

V = √ (V_R² + (V_L – Vc) ²) = √ ((IR) ² + (IX_L – IXc) ²)

V = I √ (R² + (X_L – Xc) ²)

V = IZ

Z = √ (R² + (X_L - Xc) ² )

If , V = Vm Sin wt ; i = Im Sin (wt - ф)

i.e I lags V by angle ф

Case(ii): X_L < Xc

When X_L < Xc

Also V_L < Vc (or) IX_L < IXc

Hence the resultant of V_L and Vc will directed towards Vc i.e current is said to be capacitive in nature Form voltage triangle

Diagram Fasor Aruus Rangkaian Rlc Paralel

i.e I lags V by angle ф

Case(iii): X_L = Xc

When X_L = Xc

Also V_L = Vc (or) IX_L = IXc

So V_L and Vc cancel each other and the resultant is zero. So V = V_R in such a case, the circuit is purely resistive in nature.

Impedance:

In general for RLC series circuit impedance is given by,

Z = R + j X

X = X_L – Xc = Total reactance of the circuit

Diagram Fasor Rangkaian Rlc

If XL > Xc ; X is positive & circuit is Inductive

If XL < Xc ; X is negative & circuit is Capacitive

If XL = Xc ; X =0 & circuit is purely Resistive

Tan ф = [(X_L - Xc )∕R]

Cos ф = [R∕Z]

Z = √ (R² + (X_L - Xc ) ²)

Impedance triangle:

In both cases R = Z Cos ф

X = Z Sin ф

Power and power triangle:

The average power consumed by circuit is,

Pavg = (Average power consumed by R) + (Average power consumed by L) + (Average power consumed by C)

Pavg = Power taken by R = I²R = I(IR) = VI

V = V Cos ф

P = VI Cos ф

Thus, for any condition, X_L > Xc or X_L < Xc General power can be expressed as

P = Voltage x Component in phase with voltage

Diagram Fasor Rlc Paralel

Power triangle:

S = Apparent power = I²Z = VI

P = Real or True power = VI Cos ф = Active power

Q = Reactive power = VI Sin ф