Chapter 18, Problem 115P

(a)

To determine

Find the current through the ammeter ${\text{A}}_{\text{1}}$.

Answer to Problem 115P

The current through the ammeter ${\text{A}}_{\text{1}}$ is $2.00\text{ A}$.

Explanation of Solution

Write the equation for current.

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

Here, $I$ is the current, $R_{eq}$ is the equivalent resistance and $V$ is potential difference.

Consider left top $2.00\Omega$ resistor as $R_1$, left bottom $2.00\Omega$ resistor as $R_2$, middle $2.00\Omega$ resistor as $R_3$, $3.00\Omega$ resistor as $R_4$ and $6.00\Omega$ resistor as $R_5$.

For ammeter ${\text{A}}_{\text{1}}$, the resistors $R_3$, $R_4$ and $R_5$ are in parallel then the equivalent resistance is,

$\begin{array}{l}\frac{1}{R\text{'}}=\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}\\ R\text{'}={\left(\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}\right)}^{-1}\end{array}$

This resistance is in series with $R_1$ and $R_2$, then the total equivalent resistance is,

$R_{eq}=R_1+R\text{'}+R_2$

Substitute the value of $R\text{'}$ in the above equation.

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

Conclusion:

Substitute $2.00\Omega$ for $R_1$, $2.00\Omega$ for $R_2$, $2.00\Omega$ for $R_3$, $3.00\Omega$ for $R_4$ and $6.00\Omega$ for $R_5$ in equation II.

$\begin{array}{c}{R}_{eq}=2.00\Omega +{\left(\frac{1}{2.00\Omega }+\frac{1}{3.00\Omega }+\frac{1}{6.00\Omega }\right)}^{-1}+2.00\Omega \\ =5.00\Omega \end{array}$

Substitute $5.00\Omega$ for $R_{eq}$ and $10.0\text{ V}$ for $V$ in equation I.

$\begin{array}{c}I=\frac{10.0\text{ V}}{5.00\Omega }\\ =2.00\text{ A}\end{array}$

Therefore, the current through the ammeter ${\text{A}}_{\text{1}}$ is $2.00\text{ A}$.

(b)

To determine

Find the current through the ammeter ${\text{A}}_2$.

Answer to Problem 115P

The current through the ammeter ${\text{A}}_2$ is $1.00\text{ A}$.

Explanation of Solution

Consider left top $2.00\Omega$ resistor as $R_1$, left bottom $2.00\Omega$ resistor as $R_2$, middle $2.00\Omega$ resistor as $R_3$, $3.00\Omega$ resistor as $R_4$ and $6.00\Omega$ resistor as $R_5$.

For ammeter ${\text{A}}_2$, the resistors $R_4$ and $R_5$ are in parallel then the equivalent resistance is,

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

Conclusion:

Substitute $2.00\Omega$ for $R_1$, $2.00\Omega$ for $R_2$, $2.00\Omega$ for $R_3$, $3.00\Omega$ for $R_4$ and $6.00\Omega$ for $R_5$ in the above equation.

$\begin{array}{c}R\text{'}={\left(\frac{1}{3.00\Omega }+\frac{1}{6.00\Omega }\right)}^{-1}\\ =2.00\Omega \end{array}$

Thus, the voltage across $R_3$ resistor between the ammeter is same as the $R_4$ and $R_5$ resistors as both of them has $2.00\Omega$ resistance. Then the current will spilt evenly at the first junction.

Then current through ammeter ${\text{A}}_2$ is $\frac{2.00\text{ A}}{2}=1.00\text{ A}$

Therefore, the current through the ammeter ${\text{A}}_2$ is $1.00\text{ A}$.