Chapter 18, Problem 54P

(a)

To determine

Find the equivalent resistance between point A and B.

Answer to Problem 54P

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 the resistor $2.0\Omega$ as $R_1$, top right resistor $1.0\Omega$ as $R_2$, bottom right resistor $1.0\Omega$ as $R_3$, resistor $3.0\Omega$ as $R_4$, resistor $6.0\Omega$ as $R_5$, inner left resistor $4.0\Omega$ as $R_6$, and inner right resistor $4.0\Omega$ as $R_7$.

The resistor $R_6$ and $R_7$ are in parallel then the equivalent resistance is,

$\begin{array}{l}\frac{1}{R\text{'}}=\frac{1}{R_6}+\frac{1}{R_7}\\ R\text{'}={\left(\frac{1}{R_6}+\frac{1}{R_7}\right)}^{-1}\end{array}$

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

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

This equivalent resistance is in parallel with the $R\text{'}$ equivalent resistance then the total equivalent resistance is,

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

This resistor is in series with the resistor $R_1$ and $R_4$ then the equivalent resistance is,

$R\text{'}\text{'}\text{'}\text{'}=R_1+R\text{'}\text{'}\text{'}+R_4$

This resistance is in parallel with the resistor $R_5$ then the final equivalent resistance is,

$\begin{array}{l}\frac{1}{R_{eq}}=\frac{1}{R_5}+\frac{1}{R\text{'}\text{'}\text{'}\text{'}}\\ R_{eq}=\frac{1}{\frac{1}{R_5}+\frac{1}{R\text{'}\text{'}\text{'}\text{'}}}\end{array}$

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

$R_{eq}=\frac{1}{\frac{1}{R_5}+\frac{1}{R_1+\frac{1}{\frac{1}{R_2+R_3}+\frac{1}{{\left(\frac{1}{R_6}+\frac{1}{R_7}\right)}^{-1}}}+R_4}}$

Conclusion:

Substitute $2.0\Omega$ for $R_1$, $1.0\Omega$ for $R_2$, $1.0\Omega$ for $R_3$, $3.0\Omega$ for $R_4$, $6.0\Omega$ for $R_5$, $4.0\Omega$ for $R_6$, and $4.0\Omega$ for $R_7$ in the above equation.

$\begin{array}{c}{R}_{eq}=\frac{1}{\frac{1}{6.0\Omega }+\frac{1}{2.0\Omega +\frac{1}{\frac{1}{1.0\Omega +1.0\Omega }+\frac{1}{{\left(\frac{1}{4.0\Omega }+\frac{1}{4.0\Omega }\right)}^{-1}}}+3.0\Omega }}\\ =3.0\Omega \end{array}$

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

(b)

To determine

Find the potential difference across each $4.0\Omega$ resistor.

Answer to Problem 54P

The potential difference across each $4.0\Omega$ resistor is $2.0\text{ V}$.

Explanation of Solution

The two $4.0\Omega$ resistors are parallel and series with the two $1.0\Omega$ resistors. The equivalent resistance of two $4.0\Omega$ resistor is $2.0\Omega$. The equivalent resistance of two $1.0\Omega$ resistor is $2.0\Omega$. Therefore, the equivalent resistance of four resistor is $1.0\Omega$.

This resistance is series with the $2.0\Omega$ resistor and $3.0\Omega$ resistor so their equivalent resistance is $6.0\Omega$.  This resistance is parallel with the resistor $6.0\Omega$ so then the total equivalent resistance is $3.0\Omega$ .

The current through all the resistor is $\raisebox{1ex}{$12\text{ V}$}\!\left/ \!\raisebox{-1ex}{$3.0\Omega $}\right.=4.0\text{ A}$ . This current splits equally between the equivalent resistor $6.0\Omega$  and actual resistor $6.0\Omega$.

Thus, $2.0\text{ A}$ current flows thought the $2.0\Omega$ and $3.0\Omega$ resistor and through equivalent resistance $1.0\Omega$.

Then the potential difference across the $1.0\Omega$ equivalent resistance is $\left(2.0\text{ A}\right)\left(1.0\Omega \right)=2.0\text{ V}$ .

Conclusion:

Therefore, the potential difference across each $4.0\Omega$ resistor is $2.0\text{ V}$

(c)

To determine

Find the current through the resistor of resistance $3.0\Omega$.

Answer to Problem 54P

The current through the resistor of resistance $3.0\Omega$ is $2.0\text{A}$ .

Explanation of Solution

From the discussion in part b, the current through the resistor $3.0\Omega$ is $2.0\text{A}$.

Conclusion:

Therefore, the current through the resistor of resistance $3.0\Omega$ is $2.0\text{A}$ .