Chapter 18, Problem 112P

(a)

To determine

The value of the current in ammeter ${\text{A}}_1$.

Answer to Problem 112P

The value of the current in ammeter ${\text{A}}_1$ is $\underset{\_}{1.91\text{A}}$.

Explanation of Solution

Write the expression for the current through the ammeter ${\text{A}}_1$ using ohms law.

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

Here, $I$ is the current, $V$ is the voltage, $R_{\text{eq}}$ is the equivalent resistance.

Write the expression for the equivalent resistance.

    $R_{\text{eq}}=\left[r+R_1+\frac{1}{\frac{1}{R_2}+\frac{1}{r+\left(\frac{1}{R_3}+\frac{1}{R_4}\right)}}+R_5\right]$        (II)

Here, $R_i$ is the resistance of $i^{\text{th}}$ resistor, $r$ is the resistance of ammeter.

Use equation (II) in (I) to solve for $I$.

    $I=\frac{V}{\left[r+R_1+\frac{1}{\frac{1}{R_2}+\frac{1}{r+\left(\frac{1}{R_3}+\frac{1}{R_4}\right)}}+R_5\right]}$        (III)

Conclusion:

Substitute $10\text{V}$ for $V$, $2.00\Omega$ for $R_1$, $2.00\Omega$ for $R_2$, $3.00\Omega$ for $R_3$, $6.00\Omega$ for $R_4$, $2.00\Omega$ for $R_5$, $0.200\Omega$ for $r$ in equation (III) to find $I$.

    $\begin{array}{c}I=\frac{10\text{V}}{\left[\left(0.200\Omega \right)+\left(2.00\Omega \right)+\frac{1}{\frac{1}{\left(2.00\Omega \right)}+\frac{1}{\left(0.200\Omega \right)+\left(\frac{1}{\left(3.00\Omega \right)}+\frac{1}{\left(6.00\Omega \right)}\right)}}+\left(2.00\Omega \right)\right]}\\ =\frac{10\text{V}}{4.20\Omega +\left(\frac{1}{\frac{1}{2.00\Omega }+\frac{1}{2.20\Omega }}\right)}\\ =\frac{10\text{V}}{4.20\Omega +1.048\Omega }\\ =1.905\text{A}\\ \approx 1.91\text{A}\end{array}$

Therefore, the value of the current in ammeter ${\text{A}}_1$ is $\underset{\_}{1.91\text{A}}$.

(b)

To determine

The value of the current in ammeter ${\text{A}}_2$.

Answer to Problem 112P

The value of the current in ammeter ${\text{A}}_2$ is $\underset{\_}{0.908\text{A}}$.

Explanation of Solution

Write the expression for the resistance across the branch through the ammeter ${\text{A}}_2$.

    $R=r+{\left(\frac{1}{R_3}+\frac{1}{R_4}\right)}^{-1}$        (IV)

Write the expression for the potential difference right to the ammeter ${\text{A}}_{\text{1}}$.

    $V=I\left(\frac{1}{\frac{1}{R_2}+\frac{1}{r+\left(\frac{1}{R_3}+\frac{1}{R_4}\right)}}\right)$        (V)

Write the expression for the current through the ammeter ${\text{A}}_2$.

    $I_2=VR$        (VI)

Use equation (IV) and (V) in (VI) to solve for $I_2$.

    $I_2=\frac{I\left(\frac{1}{\frac{1}{R_2}+\frac{1}{r+\left(\frac{1}{R_3}+\frac{1}{R_4}\right)}}\right)}{r+{\left(\frac{1}{R_3}+\frac{1}{R_4}\right)}^{-1}}$        (VII)

Conclusion:

Substitute $1.905\text{A}$ for $I$, $2.00\Omega$ for $R_1$, $2.00\Omega$ for $R_2$, $3.00\Omega$ for $R_3$, $6.00\Omega$ for $R_4$, $2.00\Omega$ for $R_5$, $0.200\Omega$ for $r$ in equation (VII) to find $I_2$.

    $\begin{array}{c}{I}_2=\frac{\left(1.905\text{A}\right)\left(\frac{1}{\frac{1}{\left(2.00\Omega \right)}+\frac{1}{\left(0.200\Omega \right)+\left(\frac{1}{\left(3.00\Omega \right)}+\frac{1}{\left(6.00\Omega \right)}\right)}}\right)}{\left(0.200\Omega \right)+{\left(\frac{1}{\left(3.00\Omega \right)}+\frac{1}{\left(6.00\Omega \right)}\right)}^{-1}}\\ =\frac{1.997\text{V}}{2.20\Omega }\\ =0.908\text{A}\end{array}$

Therefore, the value of the current in ammeter ${\text{A}}_2$ is $\underset{\_}{0.908\text{A}}$.