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Chapter 6

Q. 6.4

a. What is the effect of adding another resistor of 100 Ω in parallel with the parallel resistors of Example 6.1 as shown in Fig. 6.8?

b. What is the effect of adding a parallel 1 Ω resistor to the configuration in Fig. 6.8?


Verified Solution

a. Applying Eq. (6.3) gives

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)


R_T=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}=\frac{1}{\frac{1}{3\Omega }+\frac{1}{6\Omega }+\frac{1}{100\Omega } }\\[0.5cm] =\frac{1}{0.333S+0.167S+0.010S}=\frac{1}{0.510S}= 1.96\Omega

The parallel combination of the 3 Ω and 6 Ω resistors resulted in a total resistance of 2 Ω in Example 6.1. The effect of adding a resistor in parallel of 100 Ω had little effect on the total resistance because its resistance level is significantly higher (and conductance level significantly less) than that of the other two resistors. The total change in resistance was less than 2%. However, note that the total resistance dropped with the addition of the 100 Ω resistor.

b.  Applying Eq. (6.3) gives

R_T=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}}=\frac{1}{\frac{1}{3\Omega }+\frac{1}{6\Omega }+\frac{1}{100\Omega }+\frac{1}{1\Omega } }\\[0.5cm] =\frac{1}{0.333S+0.167S+0.010S+1S}=\frac{1}{0.51S}= 0.66\Omega

The introduction of the 1 Ω resistor reduced the total resistance from 2 Ω to only 0.66 Ω—a decrease of almost 67%. The fact that the added resistor has a resistance less than that of the other parallel elements and one-third that of the smallest contributed to the significant drop in resistance level.