Both batteries in Figure

(a) are ideal. Emf${\epsilon }_1$ of battery 1 has a fixed value, but emf ${\mathit{\epsilon }}_{\mathbf{1}}$of battery 2 can be varied between $\mathbf{\text{1.0V}}$and$\mathbf{\text{10V}}$ . The plots in Figure

(b) give the currents through the two batteries as a function of${\mathit{\epsilon }}_{\mathbf{2}}$ . The vertical scale is set by is${\mathit{i}}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.20}\mathbf{\text{A}}$ . You must decide which plot corresponds to which battery, but for both plots, a negative current occurs when the direction of the current through the battery is opposite the direction of that battery’s emf.

(a)What is emf${\mathit{\epsilon }}_{\mathbf{1}}$ ?

(b) What is resistance${\mathit{R}}_{\mathbf{1}}$ ?

(c) What is resistance ${\mathit{R}}_{\mathbf{2}}$?

The emf voltage,${\epsilon }_2=1.0\text{Vto}10.0\text{V}$

Vertical scale$i_s=0.20\text{A}$

Apply the Kirchhoff’s loop rule to the left-hand side loop of the circuit and use the given graph scale to find the emf ${\text{ε}}_{\text{1}}$andresistance $R_1$. Then, use these values and apply the Kirchhoff’s loop rule to the right-side loop to find the resistance$R_2$.

Formula:

The voltage, $\sum V=0$

The current, $\sum I_i=\sum I_o$

Apply the Kirchhoff’s loop rule to the left hand side loop,

${\epsilon }_2+i_1{R}_1-{\epsilon }_1=0$ ….. (1)

Since ${\epsilon }_1$has a fixed value, while ${\epsilon }_2$ and $i_1$varies.So, from the above equation, you can say that the larger values of ${\epsilon }_2$will result in a negative value for$i_1$ .

So, the downward sloping line i.e. the dashed linemust represent$i_1$.

From the graph,

The current,$i_1=0\text{A}$, when the emf voltage is${\epsilon }_2=6.0\text{V}$.

Substitute these values into equation (1).

$\begin{array}{l}6.0\text{V}+\left(0\text{A}\right)R_1-{\epsilon }_1=0\\ {\epsilon }_1=6.0\text{V}\end{array}$

Hence, the emf ${\epsilon }_1$is $6.0\text{V}$.

Again apply Kirchhoff’s loop rule to the left hand side loop.

${\epsilon }_2+i_1{R}_1-{\epsilon }_1=0$ ….. (2)

From the graph,

The current$i_1=0.20\text{A}$, when the emf voltage is${\epsilon }_2=2.0\text{V}$.

Substitute these values into equation (2).

$\begin{array}{l}2.0\text{V}+\left(0.20\text{A}\right)R_1-6.0\text{V}=0\\ R_1=\frac{6.0\text{V}-2.0\text{V}}{\left(0.20\text{A}\right)}\\ R_1=20.0\Omega \end{array}$

Hence,the resistance $R_1$ is$20.0\Omega$ .

Now, apply Kirchhoff’s loop rule to the right-hand side loop.

${\epsilon }_1-i_1{R}_1-i{R}_2=0$ ….. (3)

But, according to the junction rule,

$i=i_1+i_2$

So, by putting the above equation into equation (3), you obtain

${\epsilon }_1-i_1{R}_1-(i_1+i_2)R_2=0$ ….. (4)

From the graph,

The current, $i_2=0\text{A}$ and $i_1=0.10\text{A}$, whenthe emf voltage ${\epsilon }_2=4.0\text{V}$.

Put the known values into equation (4), and you get

$\begin{array}{l}\left(6.0\text{V}\right)-\left(0.10\text{A}\right)\left(20.0\Omega \right)-(0\text{A}+0.10\text{A})R_2=0\\ \left(6.0\text{V}\right)-\left(2.0\text{V}\right)-\left(0.10\text{A}\right)R_2=0\\ \left(4.0\text{V}\right)-\left(0.10\text{A}\right)R_2=0\end{array}$

$\begin{array}{c}{R}_2=\frac{4.0\text{V}}{0.10\text{A}}\\ =40.0\Omega \end{array}$

Hence, the resistance $R_2$ is $40.0\Omega$.