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,${\text{R}}_{\text{0}}=100\text{‰ө}$

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 a series combination of the resistors, ${\text{R}}_{\text{eq}}=\underset{\text{i}}{\overset{\text{n}}{∑}}{\text{R}}_{\text{i}}$ (2)

The equivalent resistance for the parallel combination of the resistors,

${\mathrm{R}}_{\mathrm{eq}}=\underset{\mathrm{i}}{\overset{\mathrm{n}}{∑}}\frac{1}{{\mathrm{R}}_{\mathrm{i}}}$ (3)

Since the current in the ammeter is $\mathit{i}$, the voltmeter reading is given using equation (1) as follows:

$\begin{array}{c}{\mathrm{V}}^{\text{'}}=\mathrm{V}+{\mathrm{iR}}_{\mathrm{A}}\\ =\mathrm{i}(\mathrm{R}+{\mathrm{R}}_{\mathrm{A}})\\ \mathrm{R}=\frac{{\mathrm{V}}^{\text{'}}}{\mathrm{i}}−{\mathrm{R}}_{\mathrm{A}}\\ ={\mathrm{R}}^{\text{'}}−{\mathrm{R}}_{\mathrm{A}}(âˆ\mu \text{Apparentreading,}{\mathrm{R}}^{\text{'}}=\frac{{\mathrm{V}}^{\text{'}}}{\mathrm{i}})\mathrm{............}\left(4\right)\end{array}$

Now, from the lower loop of the circuit diagram, the current through the voltmeter is given using equation (1), and the circuit resistance combination is as follows:

${\mathrm{i}}_{\mathrm{V}}=\frac{\mathrm{Î\mu }}{{\mathrm{R}}_{\mathrm{eq}}+{\mathrm{R}}_0}........................\left(5\right)$

The equivalent resistance of the circuit combination is given using equations (2) and (3) as follows:

$\begin{array}{c}\frac{1}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{1}{{\mathrm{R}}_{\mathrm{V}}}+\frac{1}{{\mathrm{R}}_{\mathrm{A}}+\mathrm{R}}\\ {\mathrm{R}}_{\mathrm{eq}}=\frac{({\mathrm{R}}_{\mathrm{A}}+\mathrm{R}){\mathrm{R}}_{\mathrm{V}}}{{\mathrm{R}}_{\mathrm{A}}+\mathrm{R}+{\mathrm{R}}_{\mathrm{V}}}\\ =\frac{(3.00\text{‰ө}+85\text{‰ө})300\text{‰ө}}{300\text{‰ө}+3.00\text{‰ө}+85\text{‰ө}}\\ =68.0\text{‰ө}\end{array}$

Then voltmeter reading is given using equation (4) and as per the problem as follows:

$\begin{array}{c}{V}^{\text{'}}=i_V{R}_{eq}\\ =\left(\frac{\mathrm{Î\mu }}{R_{eq}+R_0}\right)R_{eq}\\ =\frac{\left(12\text{ V}\right)\left(68.0\text{‰ө}\right)}{68.0\text{‰ө}+100\text{‰ө}}\\ =4.86\text{ V}\end{array}$

Thus, the ammeter reading can be given using the above value, equation (1), and the given concept as follows:

$\begin{array}{c}i=\frac{V^{\text{'}}}{R+R_A}\\ =\frac{4.86\text{ V}}{85.0\text{‰ө}+3.00\text{‰ө}}\\ =0.055\text{‼î}\end{array}$

Hence, the ammeter reading is $0.555A$.

From part (4) calculations, the voltmeter reading is found to be 4.86 V.

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

$\begin{array}{c}{\mathrm{R}}^{\text{'}}=\frac{{\mathrm{V}}^{\text{'}}}{\mathrm{i}}\\ =\frac{4.86\text{ V}}{5.52×{10}^{−2}\text{‼î}}\\ =88.0\text{‰ө}\end{array}$

Hence, the required value of the resistance is $88.0\text{‰ө}$.

From equation (4), $R=R^{\text{'}}−R_A$

If ${\mathrm{R}}_{\mathrm{A}}$ is decreased, the difference between $R\text{'}$ and $R$ decreases. Again, for the condition $R_A=0$, the relation becomes $R\text{'}=\text{R}$.

Hence, the difference value decreases.