Electrical Circuit - CSE132
- 1. ELECTRICAL CIRCUIT Course Code: CSE132
- 2. Definition : Average/active/real power, reactive power, apparent power, power factor. Relation between P & V & I Signification of Power Factor, cos θ Diagram : Phasor and Impedance Diagram Procedure to find voltage or current using Out Line:
- 3. SUBMITTED TO Mosharraf Hossain Khan (MHK) Senior Lecturer Dept. of CSESUBMITTED BY Shahriar Alam »»» 161-15-7651 Shohag Uddin »»» 163-15-8488 Md. Musfiqur Rahman Foysal »»» 163-15-8489 Amina Ahsani »»» 163-15-8533 Abdullah Bin Kamal 163-15 Tanvir Hossain »»» 163-15-8534 Shahidullah Munse »»» 163-15-8535 GROUP-4
- 4. Definition : Average power : is the power delivered by the source and diss or consumed in the load. Sometimes this is also known as Act Mathematically the equation is – P= VI cos The unit of Average power (P) is Watt. Reactive Power : The power required by a reactive load in an ac called reactive power. Reactive power is represented by (Q) Mathematically the equation is – Q = VI sin The unit of reactive power (Q) is Volt-Ampere Reactive (VAR).
- 5. Definition : Apparent power : The product of voltage and current if and on phase angle differences between current and voltage are igno Apparent power is represented by (S). Mathematically the equation is – P= VI The unit of Apparent power (S) is volt-ampere. Power Factor : the power factor of an AC electrical power syst Defined as the ratio of the real power flowing to the load to apparent power in the circuit. P= VI cos , here cos is power factor. In an AC circuit, the product of the r.m.s voltage and the r.m.s current is called apparent power.
- 6. Relation between P & V & I Let, v = Vm sin (t + v) and i = Im sin(t + i) Then power is defined by By simplifying this we get - p = VI cos(v - i) (1 - cos 2t ) + VI sin(v - i) (sin 2t) Let (v - i) = p = VI cos - VI cos (cos 2t) + VI sin (sin 2t) So average power, P = [(Vm Im/2) cos P = VI cos , this is the required equation. P = vi = Vmsin(𝜔t+𝜃v)Imsin(𝜔t+𝜃i) = VmImsin(𝜔t+𝜃v)sin(𝜔t+𝜃i) P = 𝑉 𝑚 𝐼 𝑚 2 cos(𝜃v − 𝜃i) - 𝑉 𝑚 𝐼 𝑚 2 cos(2𝜔𝑡 + 𝜃𝑣 + 𝜃𝑖) Fixed Value Time-varying (function of f)
- 7. Signification of Power Factor : Power factor = PF = cos = P/VI…… (1) The factor cos has a significant control on the delivered power By simplifying eqn (1) we get the power equation which is, P= VI cos …….(2) Here, whatever the value of V and I become if cos becomes 0 Then the power (P) becomes zero. For fig-01 P = (100 V)(5 A) (Cos 0) = 250 W For fig-02 P = (100 V)(5 A) (Cos 90) = 0 W fig-01 fig-02
- 8. Impedance Diagram Introduction :An impedance is defined when an angle is associated with resistance, inductive reactance, and capacitive reactance. each can be placed on a complex plane diagram, which is known as Impedance diagram . For any network, the resistance will always appear on the positive real axis, the inductive reactance on the positive imaginary axis, and the capacitive reactance on the negative imaginary axis
- 9. Phasor Diagram Introduction : A phasor diagram is used to show the phase relationships between two or more sine waves having the same frequency. Let, Voltage V= 20 sin ( ωt + 120 ) Current I = 5 sin ( ωt + 30 ) Here, Vm = 20 and Im = 30
- 10. Find voltage/current using Phasor Diagram i = 5sin(𝜔𝑡 + 30°) => phasor form I = 3.535 A < 30° V = IZL = (I < 𝜃)(XL< 90°) = (3.535A < 30°)(4Ω < +90°) = 14.140V < 120°) And v = 2(14.140)sin(𝜔t+120°) = 20 sin(𝜔𝑡 + 120°)
- 11. Find voltage/current using Phasor Diagram i = 15sin(𝜔𝑡 + 30°) => phasor notaion V = 10.605 V < 0° I = 𝑉 𝑍 𝑐 = 𝑉 <𝜃 𝑋 𝑐 < −90° = 10.605𝑉 <0° 2Ω <−90° = 5.303 A < 90° And i = 2(5.303)sin(𝜔t+90°) = 7.5 sin(𝜔𝑡 + 90°)
- 12. Thankyou !