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Problem 19.40 Assume that R1 48 R 78 n, Ra 20 m, R4 71 2, Rs 162, and Rs 22 2 Fi

ID: 1596042 • Letter: P

Question

Problem 19.40 Assume that R1 48 R 78 n, Ra 20 m, R4 71 2, Rs 162, and Rs 22 2 Figure 4 of 4 13 14 n O OO Part A Find the equivalent resistance of the combination shown in (Eigure 1) Express your answer in ohms to two significant figures. Ret" Submit My Answers Give U Part B Find the equivalent resistance of the combination shown in (Eigure 2) Express your answer in ohms to two significant figures. Submit My Answers Give Up Part C Find the equivalent resistance of the combination shown in (Eigure 3) Express your answer in ohms to two significant figures. Rev Submit My Answers Give Up

Explanation / Answer

if R1 and R2 are the resistances connected in series

then effective resiatcne Rnet = R1+R2

if R1 and R2 are connected in parallel,

then 1/Rnet =1/R1 +1/R2 or Rnet = R1R2/(R1+R2)


so here in fig 1:

R1 = 48 ohms

so as all are in parallel

1/Rnet = 1/25 + 1/12 + 1/5 + 1/48

1/Rnet = 0.344

Rnet in fig 1 = 3 ohms

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in figure 2,

R2 = 78 ohms

thus 78ohms and 25 ohms are in parallel

1/Rp = 1/78 +1/25

Rp = 19ohms

Now 9 ohms , 19 ohms and 18 ohms are in series

Rnet in fig 2 = Rs = 9+19+18 = 46 ohms

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in figure 3,

R3 = 20 ohms

R3 and 15 are in series, Rs = 20 +15 = 35 ohms

32ohms, 35 ohms and 14 are in parallel

1/Rp1 = 1/32 +1/35 + 1/14

Rp = 7.6 ohms

R4 = 71 ohms

R4 and 45 ohms are in parallel

1/Rp2 = 1/71 + 1/45

Rp2 = 27.54 ohms

finally

19 ohms . 7.6 ohms, 27.54 ohms and 24 ohms are in series

thus

Rnet in fig 3 = 19 + 7.6 + 27.54 + 24 = 78 ohms

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in figure 4,

R5 = 16 ohms

R6 = 22 ohms

R5, 13 ohms and 14 ohm are in Series

Rs = 16 + 13 + 14 = 43 ohms

43 ohms and R6 are in parallel

1/Rp = 1/43 +1/22

Rp = 14.55 ohms

finally 14.55 ohms is in series with 11 ohms

so

Rnet in fig 4 = 14.55 + 11 = 25.55 ohms