Chapter 18, Problem 52P

(a)

To determine

Find the equivalent resistance between point A and B.

Answer to Problem 52P

The equivalent resistance between point A and B is $3.0\Omega$.

Explanation of Solution

The sum of individual resistance connected in series and reciprocal of resistance connected in parallel will give the equivalent resistance.

From the given circuit, consider left $1.0\Omega$ as $R_1$, $2.0\Omega$ as $R_2$, top right $1.0\Omega$ as $R_3$, $4.0\Omega$ as $R_4$, $8.0\Omega$ as $R_5$, and $\text{3}\text{.3 }\Omega$ as $R_6$.

The resistor $R_4$ and $R_5$ are in parallel then the equivalent resistance is,

$\begin{array}{l}\frac{1}{R\text{'}}=\frac{1}{R_4}+\frac{1}{R_5}\\ R\text{'}={\left[\frac{1}{R_4}+\frac{1}{R_5}\right]}^{-1}\end{array}$

This resistance is in series with $R_6$ , then the equivalent resistance is,

$R\text{'}\text{'}=R_6+R\text{'}$

The resistance $R_2$ and $R_3$ are in series then the equivalent resistance is,

$R\text{'}\text{'}\text{'}=R_2+R_3$

The equivalent resistance $R\text{'}\text{'}$ and $R\text{'}\text{'}\text{'}$  are in parallel then the equivalent resistance is,

$\begin{array}{l}\frac{1}{R\text{'}\text{'}\text{'}\text{'}}=\frac{1}{R\text{'}\text{'}\text{'}}+\frac{1}{R\text{'}\text{'}}\\ R\text{'}\text{'}\text{'}\text{'}=\frac{1}{\frac{1}{R\text{'}\text{'}\text{'}}+\frac{1}{R\text{'}\text{'}}}\end{array}$

This resistance is in series with $R_1$ , then the final equivalent resistance is,

$R_{eq}=R_1+R\text{'}\text{'}\text{'}\text{'}$

Substitute the values of $R\text{'}$ , $R\text{'}\text{'}$ , $R\text{'}\text{'}\text{'}$  and $R\text{'}\text{'}\text{'}\text{'}$ in the above equation.

$R_{eq}=R_1+\frac{1}{\frac{1}{R_2+R_3}+\frac{1}{R_6+{\left[\frac{1}{R_4}+\frac{1}{R_5}\right]}^{-1}}}$

Conclusion:

Substitute $1.0\Omega$ for $R_1$, $2.0\Omega$ for $R_2$, $1.0\Omega$ for $R_3$, $4.0\Omega$ for $R_4$, $8.0\Omega$ for $R_5$, and $\text{3}\text{.3 }\Omega$ for $R_6$ in the above equation.

$\begin{array}{c}{R}_{eq}=1.0\Omega +\frac{1}{\frac{1}{2.0\Omega +1.0\Omega }+\frac{1}{3.3\Omega +{\left(\frac{1}{4.0\Omega }+\frac{1}{8.0\Omega }\right)}^{-1}}}\\ =1.0\Omega +\frac{1}{\frac{1}{2.0\Omega +1.0\Omega }+\frac{1}{3.3\Omega +2.67\Omega }}\\ =1.0\Omega +\frac{1}{\frac{1}{3.0\Omega }+\frac{1}{5.97\Omega }}\\ =1.0\Omega +1.996\Omega \\ =3.0\Omega \end{array}$

Therefore, the equivalent resistance between point A and B is $3.0\Omega$.

(b)

To determine

Find the current through the resistor of resistance $1.0\Omega$ directly connected to point A.

Answer to Problem 52P

The current through the resistor of resistance $1.0\Omega$ directly connected to point A is $6.0\text{A}$.

Explanation of Solution

Write the equation for current from emf.

$I=\frac{\epsilon }{R_{eq}}$ (I)

Here, $I$ is current, $\epsilon$ is emf and $R_{eq}$ is the equivalent resistance.

Conclusion:

Substitute $\text{18 V}$ for $\epsilon$ and $3.0\Omega$ for $R_{eq}$ in equation I.

$\begin{array}{c}I=\frac{\text{18 V}}{3.0\Omega }\\ =6.0\text{ A}\end{array}$

Therefore, the current through the resistor of resistance $1.0\Omega$ directly connected to point A is $6.0\text{A}$.

(c)

To determine

Find the current through the resistor of resistance $8.0\Omega$.

Answer to Problem 52P

The current through the resistor of resistance $8.0\Omega$ is $0.67\text{A}$ .

Explanation of Solution

The potential difference across the equivalent resistance $1.996\Omega$ for the large rectangle is $6.0\text{ A}\times 1.996\Omega =11.99\text{V}$ . The current through the bottom of large rectangle of equivalent resistance $5.97\Omega$ is $\raisebox{1ex}{$11.99\text{V}$}\!\left/ \!\raisebox{-1ex}{$5.97\Omega $}\right.=2.01\text{A}$ .

Then the potential difference across the equivalent resistance $2.67\Omega$ is $2.01\text{A}\times 2.67\Omega =5.36\text{ V}$ . The current in the resistor $8.0\Omega$ is $\raisebox{1ex}{$5.36\text{V}$}\!\left/ \!\raisebox{-1ex}{$\text{8}\text{.0 }\Omega $}\right.=0.67\text{A}$

Conclusion:

Therefore, the current through the resistor of resistance $8.0\Omega$ is $0.67\text{A}$ .