Chapter 23, Problem 97A

(a)

To determine

the resistance of the adjustable resistor.

Answer to Problem 97A

The resistance of the variable resistor is $6.0\text{kΩ}$ .

Explanation of Solution

Given:

Voltage of the battery is $V=6.0\text{V}$ .

Deflection of the ammeter is $i=1.0\text{mA}$ .

Formula used:

Resistance of the adjustable resistor is equal to the voltage applied to it divided by the current in it that is,

  $R=\frac{V}{I}$  ......(1)

Here, $V$ and $I$ are voltage of the battery and current in the variable resistor.

Calculation:

Substitute the values of $V$ and $I$ in equation (1)

  $\begin{array}{l}R=\frac{6.0\text{V}}{1.0\text{mA}}\\ R=\frac{6.0\text{V}}{1.0\times {10}^{-3}\text{A}}\\ \boxed{R=6.0\times {10}^3\Omega }\end{array}$

Conclusion:

Hence, the resistance of the variable resistor is $6.0\text{kΩ}$ .

(b)

To determine

the resistance that would be produced a current of half scale, quarter scale and three quarter scale.

Answer to Problem 97A

The resistance of the external resistor which when connected to the ohmmeter giving the half scale current, quarter scale current and three quarter scale current are $6.0\text{kΩ}$ , $18\text{kΩ}$ and $2.0\text{kΩ}$ .

Explanation of Solution

Given:

Voltage of the battery is $V=6.0\text{V}$ .

Deflection of the ammeter is $i=1.0\text{mA}$ .

Half scale current is 0.50 mA.

Quarter scale current is 0.25 mA.

Three quarter scale current is 0.25 mA.

Formula used:

In series connection of two resistors, the effective resistance is equal to the sum of the resistances of the two resistors:

  $R_{equi}=R+R_{external}$  ......(1)

Here, $R$ and $R_{external}$ are resistance of the variable resistor and external resistor.

Calculation:

But effective resistance in the circuit is given by

  $R_{effec}=\frac{V}{I}$  ......(2)

Use the expression for the expression for the effective resistance from equation (1), to replace the left hand side of this equation,

  $R+R_{external}=\frac{V}{I}$  ......(3)

Rearrange the equation (3),

  $R_{external}=\frac{V}{I}-R$  ......(4)

For half scale current,

To find resistance of the external resistor this gives the quarter scale deflection current or 0.50 mA.

Substitute the value of $V$ , $I$ and $R$ in equation (4)

  $\begin{array}{l}{R}_{external}=\frac{6\text{V}}{0.5\text{mA}}-6\times {10}^3\Omega \\ R_{external}=\frac{6\text{V}}{0.5\times {10}^{-3}\text{A}}-6\times {10}^3\Omega \\ R_{external}=12.0\times {10}^3\Omega -6.0\times {10}^3\Omega \\ R_{external}=6.0\times {10}^3\Omega \\ \boxed{R_{external}=6.0\text{k}\Omega }\end{array}$

Therefore, the resistance of the external resistor which when connected to the ohmmeter giving the half scale current is $6.0\text{kΩ}$ .

For quarter scale current,

To find resistance of the external resistor this gives the quarter scale deflection current or 0.25 mA.

Substitute the value of $V$ , $I$ and $R$ in equation (4)

  $\begin{array}{l}{R}_{external}=\frac{6\text{V}}{0.25\text{mA}}-6\times {10}^3\Omega \\ R_{external}=\frac{6\text{V}}{0.25\times {10}^{-3}\text{A}}-6\times {10}^3\Omega \\ R_{external}=24.0\times {10}^3\Omega -6.0\times {10}^3\Omega \\ R_{external}=18\times {10}^3\Omega \\ \boxed{R_{external}=18\text{k}\Omega }\end{array}$

Therefore, the resistance of the external resistor which when connected to the ohmmeter giving the quarter scale current is $18\text{kΩ}$ .

For three quarter scale current,

To find resistance of the external resistor this gives the quarter scale deflection current or 0.75 mA.

Substitute the value of $V$ , $I$ and $R$ in equation (4)

  $\begin{array}{l}{R}_{external}=\frac{6\text{V}}{0.75\text{mA}}-6\times {10}^3\Omega \\ R_{external}=\frac{6\text{V}}{0.75\times {10}^{-3}\text{A}}-6\times {10}^3\Omega \\ R_{external}=8.0\times {10}^3\Omega -6.0\times {10}^3\Omega \\ R_{external}=2.0\times {10}^3\Omega \\ \boxed{R_{external}=2.0\text{k}\Omega }\end{array}$

Therefore, the resistance of the external resistor which when connected to the ohmmeter giving the three quarter scale current is $2.0\text{kΩ}$ .

Conclusion:

Hence, the resistance of the external resistor which when connected to the ohmmeter giving the half scale current, quarter scale current and three quarter scale current are $6.0\text{kΩ}$ , $18\text{kΩ}$ and $2.0\text{kΩ}$ .