Question

Using the circuit diagram below.answer the following questions: Α Ω Β Ω 15.0 15.0V (C Ω D Ω 2T 15.0 V Assume the following resistor values for the resistors above: A = 6 Ohms B = 6 Ohms C = 8 Ohms D = 4 Ohms a) What is the value of I at resistor AQ? b) What is the potential difference across the Bo? c) What is the power at the Co? It is circled for you in the diagram.

Answer 1

Kirchhoff voltage law: The sum of all potentials in a closed circuit is zero.

Ohms law V=IR

Power P=V^2/R

Given circuit с a Ar M BR baum Yetiz) ich in 15.1 is a e mus i Dr f 15.1 By appling Kirchoff's and law to the closed loop abefa AR ال 6 b zium si Justin - A lejt is) C +15=0 Coor = 82 + + 15 V LA - é, c-igc + 1520 -ii(A+c) méąc = -15 in Captc) + igc =15 ', ( 6+8) + 128) = 15

15 = ذ 8 + رنا۱ کیاMO Now applying KVL sale to circuit "be de bi B وید الإ(ذهن) 15V Ec و ن D 0= رين+ 1 + Dون + 15-3ود + 0 = عین + c ئه م ون + 15- و ذ = 15 = 2+D+6) ون + عن 15 = (8+ 64 ) ون + (8) ذ = 2. = ون 8 + 8 کا = By solving equation ① and ② = 0. +۹۹ 4 = 0.8A 3 z A = 010A 187 ا .6

a) Current at resistor A=GR is ji - 0.8A potential difference across the BR B=64 (v=iR) Vo = 1 B x Ro VB = iq X6 VBZ 0048x6 VR22.88V 22.88V Yürtiz c C) Potential at Resistor c" Vc = (1 + i) Rc Vc - (0.8+0.48)8 VC - 10.24 V Power P = 2 2 2 (10.242 8 Pz 13-107 W