Question

In Fig. 27-76,$\mathit{R}\mathbf{=}\mathbf{10}\mathit{\Omega }$. What is the equivalent resistance between points A and B? (Hint: This circuit section might look simpler if you first assume that points A and B are connected to a battery.)

Short answer

The equivalent resistance between points A and B is $20.0\Omega$.

Step-by-step solution

The given data

The given resistance, $R=10\Omega$.

Understanding the concept of resistance

Using the concept of equivalent resistance for the series and parallel combination of resistors in the circuit, the resistance between points A and B can be calculated.

Formulae:

The equivalent resistance for a series combination,

(i)

The equivalent resistance for a parallel combination,

(ii)

Calculation of the equivalent resistance

The bottom two resistors are in parallel, equivalent to aresistance that can be given using equation (ii) as follows:

$R\text{'}=\frac{\left(6R\right)\left(3R\right)}{\left(6R\right)+\left(3R\right)}$

=2.0 R

This, then R', is in series with the resistor Ron the right so that their equivalence is given using equation (i) as follows:

$$\begin{gathered} R"=R\text{'}+R \\ =2.0R+R \\ =3.0R \end{gathered}$$

Now, near the top left are two resistors (2.0R and 4.0R) that are in series; thus, the equivalent resistance is given using equation (i) as follows:

$$\begin{gathered} R\text{'}\text{'}\text{'}=2.0R+4.0R \\ =6.0R \end{gathered}$$

Finally, R''andR'''are in parallel, so the net equivalencebetween points A and Bis given using equation (ii) as follows:

$$\begin{gathered} R_{eq}=\frac{R\text{'}\text{'}\text{'}R\text{'}}{R\text{'}\text{'}+R\text{'}} \\ =\frac{\left(3.0R\right)(6.0R)}{3.0R+6.0R} \end{gathered}$$

Substitute the values in the above expression and we get,

$$\begin{gathered} R_{eq}=2.0R \\ {R}_{eq}=2.0(10.0\Omega ) \\ =20.0\Omega \end{gathered}$$

Hence, the value of the equivalent resistance is$20.0\Omega$.