Question

Complete the required measurements and briefly explain what you observe from each circuit connection

R1 Λ- R2 ΛΜΑ- Rab R1 1 ΜΩ 100 Ω 10 kΩ R2 1 ΚΩ 100 kΩ 10 kΩ Explanation: R1 d 11 R2 gc 12 - R2 Vab Vab/ 11 12 Vde Vig Vaf Vad Vfc V R1 3.3V | 10 kΩ | 1 ΚΩ 3.3V | 1kΩ 10 ΚΩ 3V 1 ΚΩ |1kΩ Explanation:

Answer 1

Solution a R2 Ar * The glven urrit Connection is a parallel combination Rab ba I R & R2 Rab= (R: 11R2) R, R2 Ritez RM 22 = 1ks Ral (106)713) 999 on (100)+(103) RIZ = 100er, R2 = Rab = (100) (100 1103) a 99.9.2 (loe) +(100103 R = loks, R2 = 10 ks Rab = (101) (104). Do 5 km (10k) +(10 k) observation Circuit Canneets Ry & R2 in parallel don parallel combination, -Rab < minimum RiR2) R >> R₂ Rab R2) Here, Again, Ri & Ry are 5 kr с in parallel.. -9 20 I2 R2 •*; Izv RI + Iz - V R2 I e It I2

V = 3,30 -2 R1 = 10 ks; R2= 1 ks. Nab = v= 3.30 + I 1 = 1,+1 = 3,63mA I2 = v = Vie 3-3V 0:33mA RI loker 3.3 mA R2 Tkus. V de = V = 3.30 3.30 Vab I 3.3 3.63mA 909.09 Nag = ou Vadou fg = v= 3.30 V to =V=3:30 V=3.30 R1 = 1ks, R2 = 10 kr 4 WawrV = 3.3V 3.3 = 3,3ond I, 3 v Ri Iz R2 0:33 mA I = I t Iz = 3,63mA - 3:3 Hok vab I 3.3 3.63nn 909.09 videzu fg = UFC = V = 3.30 Vaf = Vadou VE 3v Ri=lker, R₂ = 1ks Vab = v=3v. I, V = 3 3 mA Ri (1 = I t Iz = 6mA) Iz | К. 3 ) и = 3M A R2 Vab I 3 0.5 km 6 ma N de = Ufg = f = 1 =3v, af = Vado Explanation Potential difference remains same Pri parallel combinatim Potential along length of a conductor is some STEP