Question: In Fig. 27-57,${\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathit{R}$, the ammeter resistance is zero, and the battery is ideal. What multiple of$\mathit{\epsilon }\mathbf{/}\mathit{R}$gives the current in the ammeter?

The given resistance value,R₁ = 2R

Ammeter resistance is zero.

When there is no voltage drop across the ammeter, then the current across each of the bottom resistors is the same and thus, the total current across the battery is the sum of current across both the resistors. And now calculating the equivalent resistance for the given circuit combination, we can get the current across each bottom resistor and also the given resistance by using Kirchhoff's voltage law. Kirchhoff’s voltage law states that in any closed loop network, the total voltage around the loop is equal to zero. Now, using all the calculated data, we can get the required ammeter reading.

Formulae:

The voltage equation using Ohm’s law, V = I R (1)

Kirchhoff’s voltage law, $\sum _{\text{closed loop}}\text{V}=0$ (2)

The equivalent resistance for series combination of the resistors, ${\text{R}}_{\text{eq}}=\sum _{\text{i}}^{\text{n}}{\text{R}}_{\text{i}}$ (3)

The equivalent resistance for parallel combination of the resistors, ${\text{R}}_{\text{eq}}=\sum _{\text{i}}^{\text{n}}\frac{\text{1}}{{\text{R}}_{\text{i}}}$ (4)

It is noted that there is no voltage drop across the ammeter. Thus, the currents in the bottom resistors are the same that is given by .

So, the current through the battery is given by and the voltage drop across each of the bottom resistors is given using equation (1) as:

The equivalent resistance of the given combination is given using equations (3) and (4) as follows:

$$\begin{gathered} R_{eq}=\frac{\left(2R\right)\left(R\right)}{2R+R}+\frac{\left(R\right)\left(R\right)}{R+R} \\ =\frac{7R}{6} \end{gathered}$$

Now, the current across the battery can be given using equation (1) as follows:

$$\begin{gathered} 2i=\frac{\epsilon }{R_{eq}} \\ i=\frac{\epsilon }{2\left(7R/6\right)} \\ i=\frac{3\epsilon }{7R} \end{gathered}$$

Using equation (2), the voltage equation for the left loop can be given as follows:

$\left({\text{Loop across the resistorsR}}_{\text{1}},\text{R}\right)$

()$$\begin{gathered} \epsilon -i_{2R}\left(2R\right)-iR=0 \\ {i}_{2R}=\frac{\epsilon -iR}{2R} \\ {i}_{2R}=\frac{\epsilon -\left(3\epsilon /7R\right)R}{2R} \\ {i}_{2R}=\frac{4\epsilon /7}{2R} \\ {i}_{2R}=\frac{2\epsilon }{7R} \end{gathered}$$

Thus, the current across the ammeter that is the ammeter reading can be given using equation (1) as follows:

$$\begin{gathered} i_{ammeter}=i-i_{2R} \\ =\frac{3\epsilon }{7R}-\frac{2\epsilon }{7R} \\ =\frac{\epsilon }{7R} \end{gathered}$$

Hence, the required value of the multiple of $\epsilon /R$is 0.143.