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Resistance of voltmeter,${\text{R}}_{\text{V}}=300\text{鈥坝�}$

Resistance of ammeter,${\text{R}}_{\text{A}}=3.00\text{鈥坝�}$

Resistance present in the circuit,$$" width="9" height="19" role="math">R0=100鈥嬧�坝�

Emf of the battery in the circuit,$蔚=12\text{鈥塚}$

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鈥檚 law,$V=IR$ (1)

The equivalent resistance for the series combination of the resistors,

${\mathrm{R}}_{\mathrm{eq}}=\underset{\mathrm{i}}{\overset{\mathrm{n}}{鈭�}}{\mathrm{R}}_{\mathrm{i}}$ (2)

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

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:

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

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

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

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

$\begin{array}{c}{i}^{\text{'}}=\frac{蔚}{R_A+R_0+\frac{R{R}_V}{R+R_V}}\\ =\frac{12.0\text{鈥塚}}{3.00\text{鈥坝�}+100\text{鈥坝�}+\left(300\text{鈥坝�}\right)\left(85\text{鈥坝�}\right)/(300\text{鈥坝�}+85\text{鈥坝�})}\end{array}$

Hence, the ammeter reading is role="math" localid="1664353694431" $7.09脳{10}^{鈭�2}\text{鈥碱}$.

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

given data as follows:

$\begin{array}{c}V=蔚鈭�i\text{'}(R_A+R_0)\\ =12.0\text{鈥塚}鈭�\left(0.0709\text{鈥碱}\right)(3.00\text{惟}+100\text{鈥嬧�坝�})\\ =4.70\text{鈥塚}\end{array}$

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

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

$\begin{array}{c}\mathrm{R}\text{'}=\frac{\mathrm{V}\text{'}}{\mathrm{i}}\\ =\frac{4.70\text{鈥塚}}{7.09脳{10}^{鈭�2}\text{A}}\\ =66.3\text{鈥坝�}\end{array}$

Hence, the required value of the resistance is $66.3\text{鈥坝�}$.

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

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

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

Hence, the difference value decreases.