Physics Homework: Reduce to a single resistor..

1). Reduce to a single resistor. Go step by step and indicate the series or parallel combinations being reduced.

Sol:

Step1: R3||R4 and R6||R7

R34 = 220*470/(220+470) = 150ohms

R67 = 330*680/(330+680) = 222.2ohms

Step2: R34(series)R2 and R67(series)R5

R2s = 150+150 = 300ohms

R5s = 100+222.2 = 322.2ohms 

Step3: [R2s||R5s](series)R1

R_eq = [300*322.2/(300+322.2)]+ 47

    = 202.35ohms

2). Reduce to a single resistor. Go step by step and indicate the series or parallel combinations being reduced.

Sol:

Step1: R3(series)R5

R35= 220+100 = 320ohms

Step2: R35||R4

R4p= 320*470/(320+470)

   = 190.38ohms

Step3: R4p(series)R1

R4p1= 190.38+47= 237.38ohms

Step4: R4p1||R2

R_eq = 237.38*150/(237.38+150)= 91.92ohms

3). Reduce to a single resistor. Go step by step and indicate the series or parallel combinations being reduced.

Sol:

Step1: R2(series)R6

R26 = 150+330= 480ohms

Step2: R26||R3

R3p = 480*220/(480+220) = 150.86ohms

Step3: R3p(series)R4

R3p4 = 150.86+470 = 620.86ohms

Step4: R3p4||R5

R5p = 620.86*100/(620.86+100)= 86.13ohms

Step5: R5p||R6

R56p = 86.13*330/(86.13+330)= 68.3ohms

Step6: R56p(series)R1

R_eq = 68.3+47 = 115.3ohms

4). Find VAB:

Sol:

Step1: R4||R5

R45 = 470*100/(470+100)= 82.46ohms

Step2: R45(series)R3

R3s = 82.46+220 = 302.46ohms

Step3: R3s||R2

R3s2 = 302.46*150/(302.46+150)= 100.27ohms

Step4: R3s2(series)R1

R1s = 100.27+47 = 147.27ohms

Total current, I=30/147.27 = 0.2037A

VAB = 0.2037*100.27 = 20.43V

5). Find VAB and the power supplied by the source:

Sol:

Step1: R3||R4

R34 = 220*470/(220+470)= 150ohms

Step2: R34(series)R2

R2s = 150+150 = 300ohms

Step3: R2s||R1

R_eq = 300*47/(300+47)= 40.63ohms

VAB = 50V

Power supplied by the source, P = VI = V²/ R_eq

                           

                                              = (50*50)/40.63

                                = 61.53W