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Chapter 6

Q. 6.2

a. By inspection, which parallel element in Fig. 6.5 has the least conductance? Determine the total conductance of the network and note whether your conclusion was verified.

b. Determine the total resistance from the results of part (a) and by applying Eq. (6.3).

R_T=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdot \cdot \cdot +\frac{1}{R_N} }                  (6.3)

Step-by-Step

Verified Solution

a. Since the 1 kΩ resistor has the largest resistance and therefore the largest opposition to the flow of charge (level of conductivity), it will have the lowest level of conductance:

G_1=\frac{1}{R_1}=\frac{1}{2\Omega }=0.5S,\, G_2=\frac{1}{R_2}=\frac{1}{200\Omega }=0.005S=5mS\\[0.5cm] G_3=\frac{1}{R_3}=\frac{1}{1k\Omega }=\frac{1}{1000\Omega }=0.001S=1mS\\[0.5cm] G_T=G_1+G_2+G_3=0.5S+5mS+1mS\\[0.5cm] =506mS

Note the difference in conductance level between the 2 Ω (500 mS) and the 1 kΩ (1 mS) resistor.

b.   R_T=\frac{1}{G_T}=\frac{1}{506mS}=1.976\Omega

Applying Eq. (6.3) gives

R_T= \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} }=\frac{1}{\frac{1}{2\Omega }+\frac{1}{200\Omega }+\frac{1}{1k\Omega } }\\[0.5cm] =\frac{1}{0.5S+0.005S+0.001S}=\frac{1}{0.506S} =1.98\Omega