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52P

Expert-verifiedFound in: Page 798

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A simple ohmmeter is made by connecting a ${\mathbf{1}}{\mathbf{.50}}{\mathbf{\text{\hspace{0.17em}V}}}$**** flashlight battery in series with a resistance ${\mathbf{\text{R}}}$**** and an ammeter that reads from 0 to ${\mathbf{1.00}}{\mathbf{\text{\hspace{0.17em}mA}}}$****, as shown in Fig. 27-59. Resistance ${\mathbf{\text{R}}}$**** is adjusted so that when the clip leads are shorted together, the meter deflects to its full-scale value of ${\mathbf{1}}{\mathbf{.00}}{\mathbf{}}{\mathbf{\text{mA}}}$****. What external resistance across the leads results in a deflection of (a)** ${\mathbf{10.0}}{\mathit{\%}}$**, (b)** ${\mathbf{50.0}}{\mathit{\%}}$**, and (c)** ** ${\mathbf{90.0}}{\mathit{\%}}$of full scale? (d) If the ammeter has a resistance of**** ${\mathbf{20.0}}{\mathbf{}}{\mathbf{\text{\Omega}}}$and the internal resistance of the battery is negligible, what is the value of ${\mathbf{\text{R}}}$****?**

- The external resistance that results across the leads results in a deflection of $10\%$ is $1.35\times {10}^{4}\text{\hspace{0.17em}\Omega}$.
- The external resistance that results across the leads results in a deflection of $50\%$ is $1.5\times {10}^{3}\text{\hspace{0.17em}\Omega}$.
- The external resistance that results across the leads results in a deflection of $90.0\%$ full scale is $167\text{\hspace{0.17em}\Omega}$.
- The value of $R$ is $1.48\times {10}^{3}\text{\hspace{0.17em}\Omega}$.

The voltage of the flashlight battery, $\text{V}=1.50\text{\hspace{0.17em}V}$

Ammeter reads from $0\text{\hspace{0.17em}}\mathrm{mA}$ to $1.00\text{\hspace{0.17em}mA}$

Deflection of ammeter after adjustments of the resistance, $\text{i}=1.00\text{\hspace{0.17em}mA}$

Resistance of ammeter, $\text{R}=20\text{\hspace{0.17em}\Omega}$

**In a given figure, we need to check the resistance combination of the circuit, to gain a clear idea of the equivalent resistance due to the contribution of every resistance present in the circuit. Now, using Ohm's law in the given situation, we can solve for the maximum and variable current case to get the required resistances for each case.**

Formulae:

The voltage equation using Ohm’s law, $V=IR$ (1)

The equivalent resistance for a series combination of the resistors, ${\mathrm{R}}_{\mathrm{eq}}=\sum _{\mathrm{i}}^{\mathrm{n}}{\mathrm{R}}_{\mathrm{i}}$ (2)

From the given combination, the current flowing through the circuit can be given using equation (2) in equation (1) as follows:

$\begin{array}{c}\mathrm{i}=\frac{\mathrm{\epsilon}}{\mathrm{R}+{\mathrm{R}}_{\mathrm{ext}}}\\ \mathrm{i}=\frac{{\mathrm{Ri}}_{\mathrm{max}}}{\mathrm{R}+{\mathrm{R}}_{\mathrm{ext}}}\text{}\left(\because {\mathrm{i}}_{\mathrm{max}}=\frac{\mathrm{\epsilon}}{\mathrm{R}}\right)\\ {\mathrm{R}}_{\mathrm{ext}}=\mathrm{R}\left(\frac{{\mathrm{i}}_{\mathrm{max}}}{\mathrm{i}}-1\right)\mathrm{........................}\left(3\right)\end{array}$

Now, the internal resistance can be given using equation (1) as follows:

$$" width="9" height="19" role="math" style="max-width: none; vertical-align: -4px;" localid="1664352318254">$$" width="9" height="19" role="math" style="max-width: none; vertical-align: -4px;" localid="1664352323339">

$\begin{array}{c}R=\frac{\epsilon}{{i}_{\mathrm{max}}}\\ =\frac{1.50\text{\hspace{0.17em}V}}{1.00\times {10}^{-3}\text{A}}\\ =1.50\times {10}^{3}\text{\hspace{0.17em}\Omega}\end{array}$Thus, the external resistance value for the current value getting 10% deflection is given using the given data in equation (3) as follows:

$\begin{array}{c}{R}_{ext}=1.50\times {10}^{3}\text{\hspace{0.17em}\Omega}\left(\frac{1}{0.10}-1\right)\\ =1.35\times {10}^{4}\text{\hspace{0.17em}\Omega}\end{array}$

Hence, the value of the external resistance is $1.35\times {10}^{4}\text{\hspace{0.17em}\Omega}$.

The external resistance value for the current value getting 50% deflection is given using the given data in equation (3) as follows:

$\begin{array}{c}{\mathrm{R}}_{\mathrm{ext}}=1.50\times {10}^{3}\text{\hspace{0.17em}\Omega}\left(\frac{1}{0.50}-1\right)\\ =1.5\times {10}^{3}\text{\hspace{0.17em}\Omega}\end{array}$

Hence, the value of the external resistance is $1.5\times {10}^{3}\text{\hspace{0.17em}\Omega}$ .

The external resistance value for the current value getting 90% deflection is given using the given data in equation (a) as follows:

$\begin{array}{c}{\mathrm{R}}_{\mathrm{ext}}=1.50\times {10}^{3}\text{\hspace{0.17em}\Omega}\left(\frac{1}{0.90}-1\right)\\ =167\text{\hspace{0.17em}\Omega}\end{array}$

Hence, the value of the external resistance is $167\text{\hspace{0.17em}\Omega}$.

With the ammeter having a resistance value, the total internal resistance using equation (2) for the given circuit becomes:

$\begin{array}{l}\mathrm{r}={\mathrm{R}}_{\mathrm{A}}+\mathrm{R}\text{}(\mathrm{where},\text{r is the resitance of the flashlight battery})\\ \mathrm{R}=\mathrm{r}-{\mathrm{R}}_{\mathrm{A}}\\ \mathrm{R}=\frac{\mathrm{\epsilon}}{{\mathrm{i}}_{\mathrm{max}}}-{\mathrm{R}}_{\mathrm{A}}\\ \mathrm{R}=\frac{1.50\text{\hspace{0.17em}V}}{1.00\times {10}^{-3}\text{\hspace{0.17em}A}}-20\text{\hspace{0.17em}\Omega}\\ \mathrm{R}=1500\text{\hspace{0.17em}\Omega}-20\text{\hspace{0.17em}\Omega}\\ \mathrm{R}=1.48\times {10}^{3}\text{\hspace{0.17em}\Omega}\end{array}$

Hence, the value of the unknown resistance for the given ammeter reading is $1.48\times {10}^{3}\text{\hspace{0.17em}\Omega}$.

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