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. 6.2

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) ## 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

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