In Fig. 27-62, a voltmeter of resistance ${\mathbf{R}}_{\mathbf{V}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{300}\mathbf{}\mathbf{\text{Ω}}$and an ammeter of resistance ${\mathbf{R}}_{\mathbf{A}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3.00}\mathbf{}\mathbf{\text{Ω}}$are being used to measure a resistance $\mathbf{R}$in a circuit that also contains a resistance ${\mathbf{R}}_{\mathbf{0}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{100}\mathbf{}\mathbf{\text{Ω}}$and an ideal battery of emf role="math" localid="1664352839658" $\mathbf{\epsilon }\mathbf{}\mathbf{=}\mathbf{}\mathbf{12.0}\mathbf{}\mathbf{\text{V}}$. Resistance $\mathbf{\text{R}}$is given by$\mathbf{R}\mathbf{}\mathbf{=}\mathbf{}\mathbf{V}\mathbf{/}\mathbf{i}$ , where V is the voltmeter reading and is the current in resistance $\mathbf{R}\mathbf{}$. However, the ammeter reading is $\mathbf{i}\mathbf{}$not but rather $\mathbf{i}\mathbf{\text{'}}\mathbf{}$, which is $\mathbf{i}\mathbf{}$plus the current through the voltmeter. Thus, the ratio of the two meter readings is not$\mathbf{R}$ but only an apparent resistance$\mathbf{R}\mathbf{\text{'}}\mathbf{=}\mathbf{}\mathbf{V}\mathbf{/}\mathbf{i}\mathbf{\text{'}}$ . If$\mathbf{R}\mathbf{}\mathbf{=}\mathbf{}\mathbf{85.0}\mathbf{}\mathbf{\text{Ω}}$ , what are (a) the ammeter reading, (b) the voltmeter reading, and (c) $\mathbf{R}\mathbf{}\mathbf{\text{'}}$? (d) If${\mathbf{R}}_{\mathbf{V}}$ is increased, does the difference between $\mathbf{R}{\mathbf{\text{'}}}_{}$and $\mathbf{R}$increase, decrease, or remain the same?

Resistance of voltmeter,${\text{R}}_{\text{V}}=300\text{\hspace{0.17em}Ω}$

Resistance of ammeter,${\text{R}}_{\text{A}}=3.00\text{\hspace{0.17em}Ω}$

Resistance present in the circuit,$$" width="9" height="19" role="math">R0=100​ Ω

Emf of the battery in the circuit,$\epsilon =12\text{\hspace{0.17em}V}$

Extra resistance given is $\text{R'}$ and $\text{R}$.

Ammeters measure the current of a circuit, and voltmeters measure the voltage drop across a resistor. Thus, using this concept and the use of Ohm's law will determine the required unknown parameters.

Formulae:

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

The equivalent resistance for the series combination of the resistors,

${\mathrm{R}}_{\mathrm{eq}}=\sum _{\mathrm{i}}^{\mathrm{n}}{\mathrm{R}}_{\mathrm{i}}$ (2)

The currents in $\text{R}$ and ${\text{R}}_{\text{V}}$ are $\text{i}$ and $(i-\text{i})$.

Now, the voltmeter reading is the same for the parallel connection of $\text{R}$ and ${\text{R}}_{\text{V}}$ , thus using equation (1) we get the following resistance relation:

$\begin{array}{c}\mathrm{V}=\mathrm{iR}\\ \mathrm{V}=(\mathrm{i}\text{'}-\mathrm{i}){\mathrm{R}}_{\mathrm{V}}\\ 1=\left(\frac{\mathrm{i}\text{'}}{\mathrm{V}}-\frac{\mathrm{i}}{\mathrm{V}}\right){\mathrm{R}}_{\mathrm{V}}\\ \frac{1}{{\mathrm{R}}_{\mathrm{V}}}=\frac{1}{\mathrm{R}\text{'}}-\frac{1}{\mathrm{R}}\\ \frac{1}{\mathrm{R}\text{'}}=\frac{1}{{\mathrm{R}}_{\mathrm{V}}}+\frac{1}{\mathrm{R}}\\ \mathrm{R}\text{'}=\frac{{\mathrm{RR}}_{\mathrm{V}}}{\mathrm{R}+{\mathrm{R}}_{\mathrm{V}}}...............................\left(3\right)\end{array}$

Now, the equivalent resistance of the circuit combination is given using equations (3) as follows:

$\begin{array}{c}{R}_{eq}=R_A+R_0+R\text{'}\\ =R_A+R_0+\frac{R{R}_V}{R+R_V}(\because \text{Fromequation(3)})\end{array}$

Thus, the ammeter reading can be given using the above value, equation (3), and the given concept as follows:

$\begin{array}{c}{i}^{\text{'}}=\frac{\epsilon }{R_A+R_0+\frac{R{R}_V}{R+R_V}}\\ =\frac{12.0\text{\hspace{0.17em}V}}{3.00\text{\hspace{0.17em}Ω}+100\text{\hspace{0.17em}Ω}+\left(300\text{\hspace{0.17em}Ω}\right)\left(85\text{\hspace{0.17em}Ω}\right)/(300\text{\hspace{0.17em}Ω}+85\text{\hspace{0.17em}Ω})}\end{array}$

Hence, the ammeter reading is role="math" localid="1664353694431" $7.09\times {10}^{-2}\text{\hspace{0.17em}A}$.

Now, the voltmeter reading is given by using equation (1) in equation (2) with the

given data as follows:

$\begin{array}{c}V=\epsilon -i\text{'}(R_A+R_0)\\ =12.0\text{\hspace{0.17em}V}-\left(0.0709\text{\hspace{0.17em}A}\right)(3.00\text{Ω}+100\text{\hspace{0.17em}Ω})\\ =4.70\text{\hspace{0.17em}V}\end{array}$

Hence, the voltmeter reading is $4.70\text{\hspace{0.17em}V}$.

As per the problem, the value of the resistance is given as follows:

$\begin{array}{c}\mathrm{R}\text{'}=\frac{\mathrm{V}\text{'}}{\mathrm{i}}\\ =\frac{4.70\text{\hspace{0.17em}V}}{7.09\times {10}^{-2}\text{A}}\\ =66.3\text{\hspace{0.17em}Ω}\end{array}$

Hence, the required value of the resistance is $66.3\text{\hspace{0.17em}Ω}$.

From equation (3), the resistance relation can be given as:

$\begin{array}{c}R\text{'}-R=\frac{R{R}_V}{R+R_V}-R\\ =\frac{R{R}_V-R^2-R{R}_V}{R+R_V}\\ =\frac{-R^2}{R+R_V}\end{array}$

If $R_A$ is decreased, the difference between $\text{R'}$ and $\text{R}$ decreases from the above relation. Again, for the condition ${\text{R}}_{\text{V}}\to \infty$, the relation becomes $\text{R'}\to \text{R}$.

Hence, the difference value decreases.