Chapter 28: Q. 28 (page 791)

What is the equivalent resistance between points a and b in FIGURE EX28.28?

Short Answer

Expert verified

The equivalent resistance between points a and b is $R_{a b} = 54.5 Ω$

Step by step solution

Given Information

We need to find the equivalent resistance between points a and b.

Simplify

The resistors are connected one by one and with no junctions between them then they are called to be connected in series. For a series resistors, the equivalent resistance for their combination is given by equation in the form

$R_{e q} = R_{1} + R_{2} + . . . . + R_{N}$ (1)

In figure EX28.28, there are three resistors , and are connected in series, so let us find the equivalent resistance for this combination using equation (1) by

$R_{1, e q} = 100 Ω + 100 Ω + 100 Ω = 300 Ω$

Also, there are two resistors that are connected in series. Calculating the equivalent resistance for this combination using equation (1) by

$R_{2, e q} = 100 Ω + 100 Ω = 200 Ω$

Between the two points a and b, the two combinations $R_{2, e q}$ and $R_{1, e q}$ are connected in parallel with each other and with the resistance $100 Ω$ . The equation shows the equivalent resistance for the parallel connection in the form

$\frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + . . . . + \frac{1}{R_{N}}$ (2)

We use the values for $R_{2, e q,} R_{1, e q}$ , and $100 Ω$ into equation (2) to get the equivalent resistance between points a and b by

$\frac{1}{R_{a b}} = \frac{1}{R_{1, e q}} + \frac{1}{R_{2, e q}} + \frac{1}{100 Ω} \frac{1}{R_{a b}} = \frac{1}{200 Ω} + \frac{1}{300 Ω} + \frac{1}{100 Ω} R_{a b} = {(\frac{1}{200 Ω} + \frac{1}{300 Ω} + \frac{1}{100 Ω})}^{- 1} R_{a b} = 54.5 Ω$