Question

R2=100 Part II: 1. Series connection - R = 52 1. Output of power supply = 5V 3. Vi= 1.227 v . 4. V2= 2.460 5. Vz+V2= 3.687 Comment : 6. I (before Ri) = 0.23. , I (between R1 & R2)= 0.234_,I(afterRz)= 0.23A Comment: 7. Compare the product Rix I to V1. 8. Compare the product RzxI to V2. b) Parallel connection R-5.12 Rozloc 2. Output of power supply 5V 3. V₂ = 2,939 v 4. V2 = 2.226 v vt= 5. Comment 6. I1 = 0.55A 7. I2 = 0.27 8. It = 0.83 4 9. 11 + 12 = 10. Comment Conclusions:

What is the conclusion needed

Answer 1

Conclusions 1) There in some internal scointanu & power supply an Vtvo = 3.6874 075V Coerico de bedre voksne e perdella Also an IXR = 0.23 X5 2 1-15 2) and y = 1.227V so of similarity IXR₂ 2 0.23x102 2.3v. & V₂ = 2.46 v there in some internal reontane Ammeter also, in calculating the serico con connection of Ry & ka 3 Also in porallel case there is some internal resistance is power supply as Inet = 5 + 5 (v + 3 2 1. 5A but it in 0.83 A bet 4) In serico combination current remain came l.p. 0.23 A access the resistance In parallel combination voltage remain some ie . n 2.930 acroom R, BR All error occury due to boternal recintanu Bower supply and multimeter 6)