kumar ideology

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...

A (50/60 Hz) Transformer. Which one will give more Output? (When operates on 50 or 60 Hz frequency) A Transformer is designed to be operated on both 50 & 60 Hz frequency.  For the Same rating, which one will give more out put; when,   Operates on 50 Hz    Operates on 60 Hz Obviously!  It will give more out put when we operate a transformer (of same rating) on 50 Hz instead of 60 Hz. Because in previous posts, we proved that, in inductive circuit,  when frequency increases, the circuit power factor decreases . Consequently, the transformer out put decreases. Let’s consider the following example. Suppose, When Transformer operates on  50 Hz  Frequency Transformer = 100kVA, R=700 Ω , L=1.2 H, f= 50 Hz. X L  = 2πfL = 2 x 3.1415 x 50 x 1.2 = 377 Ω Impedance Z = √ (R 2 +X L 2 ) = √ (700 2 + 377 2 ) = 795 Ω Power factor Cos θ = R/Z = 700/795 = 0.88 Transformer Output (Real Power) kVA x Cos θ 100kVA...

Why we can’t store AC in Batteries instead of DC.or Can we store AC in batteries instead of DC? We  cannot store AC in batteries  because AC changes their polarity up to 50 (When frequency = 50 Hz) or 60 (When frequency = 60 Hz) times in a second. Therefore the battery terminals keep changing Positive (+ve) becomes Negative (-Ve) and vice versa, but the battery cannot change their terminals with the same speed so that’s why we can’t store AC in Batteries. in addition, when we  connect a battery with AC Supply , then It will charge during positive half cycle and discharge during negative half cycle, because the Positive (+ve) half cycle cancel the negative (-Ve) half cycle, so the  average voltage or current in a complete cycle is Zero . So there is no chance to store AC in the Batteries. If you want more books, slide, ppt's and notes,Request in the comment section. Visit the site for more knowledge and share this site link to the needy one....