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What is the Role of Capacitor in AC and DC Circuit? Role of Capacitor in AC Circuits:  In an AC circuit, capacitor reverses its charges as the current alternates and produces a lagging voltage (in other words, capacitor provides leading current in AC circuits and networks)  Role and Performance of Capacitor in DC Circuit In a DC Circuit, the capacitor once charged with the applied voltage acts as an open switch. What is the Role of Capacitor in AC and DC Circuit? Let’s explain in detail, but we will go back to the basics of capacitor first to discuss the matter. What is Capacitor? The capacitor is a two terminal electrical device used to store electrical energy in the form of electric field between the two plates. It is also known as a condenser and the SI unit of its capacitance measure is Farad “F”, where Farad is a large unit of capacitance, so they are using microfarads (µF) or nanofarads (nF) nowadays. How Capacitor Works? Worki...

How To remember the direction of PNP and NPN Transistor. One of the Easiest way. How To remember the direction of PNP and NPN Transistor. One of the Easiest way. The Easiest way to remember the direction of PNP and NPN Transistor. PNP = Pointed In NPN = Not Pointed In. if you think that is little bit complex, then try this one..it’s more simple. PNP                                 NPN P  =  P oints                       N  =  N ever N  = i N                              P  =  P oints P  =  P ermanently   ...

Easy Battery Charging Time and battery Charging Current Formula for Batteries. ( with Example of 120Ah Battery). How to Calculate the charging Time and Charging current for battery Charging? Easy Battery Charging Time and battery Charging Current Formula for Batteries. ( with Example of 120Ah Battery). Battery Charging Current and Battery Charging Time formula Here is the formula of Charging Time of a Lead acid battery.  Charging Time of battery = Battery Ah / Charging Current   T = Ah / A   Example, Suppose for 120 Ah battery,  First of all, we will calculate charging current for 120 Ah battery. As we know that charging current should be 10% of the Ah rating of battery. so charging current for120Ah Battery = 120 x (10/100) = 12 Amperes. but due to losses, we can take 12-14Amperes for charging purpose. suppose we took 13 Amp for charging purpose,  then charging time for 120Ah battery = 120 / 13 = 9.23 Hrs. but this was an ideal ...

What is Main Difference b/w Electrical and Electronics Engineering? What is Main Difference b/w Electrical and Electronics Engineering? According to Wikipedia; Electrical Engineering  is the field of Engineering that generally deals with the study and application of electricity, electronics, and electromagnetism. and, Electronics Engineering   is an Engineering discipline where non-linear and active electrical and electronics components and devices such as electron tubes, and semiconductor devices, especially transistors, diodes and integrated circuits, etc. are utilized to design electronic circuits, devices and systems. but below is the main difference between Electrical and Electronics Engineering which prevents such a confusion between Electrical and Electronics Engineering Difference between Electrical and Electronics Engineering. Main difference between Electrical and Electronics Engineering. Electrical Engineering  = Study and Utiliz...

Why the Circuit Power factor (Cos θ) Decreases, when Inductance (L) or inductive reactance (XL) increases, In inductive circuit? OR Explain the statement that  “  the Circuit Power factor   (Cos θ)  Decreases,  when Inductance (L) or inductive reactance (X L ) increases” Explanation:    Suppose,  when Inductance (L) = 0.02H V=220, R= 10 Ω,  L=0.02 H , f=50Hz. X L  = 2πfL = 2 x 3.1415 x 50 x 0.02 = 6.28 Ω Z = √ (R 2 +X L 2 ) = √ (10 2  + 6.28 2 ) = 11.8 Ω Cos θ = R/Z = 10/11.05 =   0.85 Now we increases Inductance (L) form 0.02 H to 0.04 H, V=220, R= 10 Ω,  L=0.04 H , f=50Hz. X L  = 2πfL= 2 x 3.1415 x 50 x 0.04 = 12.56 Ω Z = √ (R 2 +X L 2 ) = √ (10 2  + 12.56 2 ) = 16.05 Ω Cos θ = R/Z = 10/16.05 =  0.75 Conclusion : We can see that, When inductance (L) was 0.02, then circuit current were 18.64 A, and Circuit power factor was (Co...

Why the circuit Current (I) decrease, when Inductance (L) or inductive reactance (XL) increases in inductive circuit?   Explain the statement that ” In Inductive circuit, when  Inductance (L) or inductive reactance (X L ) increases, the circuit Current (I) decrease”   OR Why the circuit Current (I) decrease, when   Inductance (L) or inductive reactance (X L ) increases in inductive circuit? Explanation: We know that,  I = V / R,  but in inductive circuit,  I = V/X L So Current in inversely proportional  to the Current ( in inductive circuit. Let ‘s check with an example..     Suppose, when Inductance (L) = 0.02H V=220, R= 10 Ω,  L=0.02 H , f=50Hz. X L  = 2πfL = 2 x 3.1415 x 50 x 0.02 = 6.28 Ω Z = √ (R 2 +X L 2 ) = √ (10 2  + 6.28 2 ) = 11.8 Ω I = V/Z = 220/11.8 =   18.64 A Now we increases Inductance (L) form 0.02 H to 0.04 H, V=2...