Chapter 18, Problem 116P

(a)

To determine

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

Answer to Problem 116P

The current through the ammeter ${\text{A}}_{\text{1}}$ is $\text{1}\text{.91 A}$.

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 $R_4$, $6.00\Omega$ resistor as $R_5$, resistance of ammeter ${\text{A}}_{\text{1}}$ as $R_6$ and resistance of ammeter ${\text{A}}_2$ as $R_7$

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

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

This resistance is in series with $R_7$, then the equivalent resistance is,

$R\text{'}\text{'}=R_7+R\text{'}$

This resistance is parallel with $R_3$, then the equivalent resistance is,

$\begin{array}{l}\frac{1}{R\text{'}\text{'}\text{'}}=\frac{1}{R_3}+\frac{1}{R\text{'}\text{'}}\\ R\text{'}\text{'}\text{'}=\frac{1}{\frac{1}{R_3}+\frac{1}{R\text{'}\text{'}}}\end{array}$

This resistance is in series with $R_1$, $R_6$ and $R_2$, then the final equivalent resistance is,

$R_{eq}=R_1+R_6+R\text{'}\text{'}\text{'}+R{2}_{}$

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

$R_{eq}=R_1+R_6+\frac{1}{\frac{1}{R_3}+\frac{1}{R_7+{\left(\frac{1}{R_4}+\frac{1}{R_5}\right)}^{-1}}}+R_2$ (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 $R_4$, $6.00\Omega$ for $R_5$, $0.200\Omega$ for $R_6$ and $0.200\Omega$ for $R_7$ in equation II.

$\begin{array}{c}{R}_{eq}=2.00\Omega +0.200\Omega +\frac{1}{\frac{1}{2.00\Omega }+\frac{1}{0.200\Omega +{\left(\frac{1}{3.00\Omega }+\frac{1}{6.00\Omega }\right)}^{-1}}}+2.00\Omega \\ =4.20\Omega +1.048\Omega \\ =5.248\Omega \end{array}$

Substitute $5.248\Omega$ for $R_{eq}$ and $10.0\text{ V}$ for $V$ in equation I.

$\begin{array}{c}I=\frac{10.0\text{ V}}{5.248\Omega }\\ =1.905\text{ A}\\ \simeq 1.91\text{ A}\end{array}$

Therefore, the current through the ammeter ${\text{A}}_{\text{1}}$ is $\text{1}\text{.91 A}$.

(b)

To determine

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

Answer to Problem 116P

The current through the ammeter ${\text{A}}_2$ is $0.908\text{ A}$.

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 $R_4$, $6.00\Omega$ resistor as $R_5$, resistance of ammeter ${\text{A}}_{\text{1}}$ as $R_6$ and resistance of ammeter ${\text{A}}_2$ as $R_7$

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

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

This resistor is in series with $R_7$, then the effective resistance is,

$R_{ef}=R_7+R\text{'}$

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

$R_{ef}=R_7+{\left(\frac{1}{R_4}+\frac{1}{R_5}\right)}^{-1}$

Conclusion:

Substitute $2.00\Omega$ for $R_1$, $2.00\Omega$ for $R_2$, $2.00\Omega$ for $R_3$, $3.00\Omega$ for $R_4$ $6.00\Omega$ for $R_5$, $0.200\Omega$ for $R_6$ and $0.200\Omega$ for $R_7$ in above equation.

$\begin{array}{c}{R}_{ef}=0.200\Omega +{\left(\frac{1}{3.00\Omega }+\frac{1}{6.00\Omega }\right)}^{-1}\\ =2.20\Omega \end{array}$

From the previous part the effective resistance right of ammeter ${\text{A}}_1$ is $1.048\Omega$. Then the potential difference across this resistance is,

$\left(1.905\text{ A}\right)\left(1.048\Omega \right)=1.997\text{ V}$

This is the voltage across the ammeter ${\text{A}}_2$.

Then current through ammeter ${\text{A}}_2$ is $\frac{1.997\text{ V}}{2.20\Omega }=0.908\text{ A}$

Therefore, the current through the ammeter ${\text{A}}_2$ is $0.908\text{ A}$.