Given the following, calculate the total conductance (or resistance as specified

Question

Given the following, calculate the total conductance (or resistance as specified) in the following configurations. Draw an element combination diagram of the pathways in each case (combinations of elements as we discussed in class).

g1 = 1 mol m-2 s -1

g2 = 2 mol m-2 s -1

g3 = 3 mol m-2 s -1

r4 = 0.25 m 2 s mol-1 (note the last one is specified as a resistance)

a) the conductance of g1 and g2 in series

b) the conductance of g1, g2, g3, and g4 in series

c) the resistance of g1, g2, g3, and g4 in parallel

d) the conductance of the combination of g1 and g2 in parallel

e) the conductance of (g1 and g2 in series) in series with (g3 and g4 in parallel)

For c I got 2.083 by adding all 1/g1+1/g2+1/g3+r4

Explanation / Answer

a)

In series combination of g1 and g2 is given as

1/g12 = 1/g1 + 1/g2

1/g12 = 1/1 + 1/2

g12 = 0.67

b)

In series combination of g1, g2, g3, and g4 is given as

1/g1234 = 1/g1 + 1/g2 + 1/g3 + 1/g4

1/g1234 = 1/1 + 1/2 + 1/3 + 4

g1234 = 0.48

c)

1/R1234 = 1/R1 + 1/R2 + 1/R3 + 1/R4

1/R1234 = g1 + g2 + g3 + g4

1/R1234 = 1 + 2 + 3 + 0.25

R1234 = 0.16

d)

1/R12 = 1/R1 + 1/R2

g12 = g1 + g2

g12 = 1 + 2 = 3

e)

R12 = R1 + R2 = 1/g1 + 1/g2 = 1/1 + 1/2 = 1.5

1/R34 = 1/R3 + 1/R4 = 1/3 +0.25

R34 = 1.71

R1234 = R12 + R34 = 1.5 + 1.71 = 3.21

g1234 = 1/R1234 = 1/3.21 = 0.312

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