91Ó°ÊÓ

Resistance of voltmeter,${\text{R}}_{\text{V}}=300\text{‰ө}$

Resistance of ammeter,${\text{R}}_{\text{A}}=3.00\text{‰ө}$

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

Emf of the battery in the circuit,$\mathrm{Î\mu }=12\text{ V}$

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,

${\mathrm{R}}_{\mathrm{eq}}=\underset{\mathrm{i}}{\overset{\mathrm{n}}{∑}}{\mathrm{R}}_{\mathrm{i}}$ (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:

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

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

$\begin{array}{c}{i}^{\text{'}}=\frac{\mathrm{Î\mu }}{R_A+R_0+\frac{R{R}_V}{R+R_V}}\\ =\frac{12.0\text{ V}}{3.00\text{‰ө}+100\text{‰ө}+\left(300\text{‰ө}\right)\left(85\text{‰ө}\right)/(300\text{‰ө}+85\text{‰ө})}\end{array}$

Hence, the ammeter reading is role="math" localid="1664353694431" $7.09×{10}^{−2}\text{‼î}$.

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

given data as follows:

$\begin{array}{c}V=\mathrm{Î\mu }−i\text{'}(R_A+R_0)\\ =12.0\text{ V}−\left(0.0709\text{‼î}\right)(3.00\text{Ω}+100\text{​‰ө})\\ =4.70\text{ V}\end{array}$

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

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

$\begin{array}{c}\mathrm{R}\text{'}=\frac{\mathrm{V}\text{'}}{\mathrm{i}}\\ =\frac{4.70\text{ V}}{7.09×{10}^{−2}\text{A}}\\ =66.3\text{‰ө}\end{array}$

Hence, the required value of the resistance is $66.3\text{‰ө}$.

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 ${\text{R}}_{\text{V}}→\mathrm{∞}$, the relation becomes $\text{R'}→\text{R}$.

Hence, the difference value decreases.