Products
Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY
HELP DESK

Need Help? We got you covered.

## 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.