(a) What is the potential difference ${\mathbf{V}}_{\mathbf{ad}}$in the circuit of Fig. P25.62? (b) What is the terminal voltage of the $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{-}\mathbf{V}$battery? (c) A battery with emf and internal resistance $\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\Omega }$is inserted in the circuit at d, with its negative terminal connected to the negative terminal of the $\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{-}\mathbf{V}$battery. What is the difference of potential ${\mathbf{V}}_{\mathbf{bc}}$between the terminals of the $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{-}\mathbf{V}$battery now?

Consider the expression for the ohm’s law.

$V=IR$

Here,$I$ is current in ampere $\left(A\right)$, $R$is resistance in ohms$\left(\mathrm{\Omega }\right)$ and$V$ is the potential difference volt $\left(V\right)$.

Consider the real source of emf $\epsilon$has internal energy $r$, then its terminal potential difference is as follows:

${\mathbf{V}}_{\mathbf{ab}}\mathbf{=}\mathbf{\epsilon }\mathbf{‐}\mathbf{IR}$

(a)

Given the value of the internal resistance, load resistance and the battery emf as follows:

$$\begin{gathered} r=0.5\mathrm{\Omega } \\ R=8.0\mathrm{\Omega } \\ \epsilon =8.0\mathrm{V} \end{gathered}$$

From the ohm’s law determine the current in the circuit.

$$\begin{gathered} I=\frac{8\mathrm{V}-4\mathrm{V}}{8\mathrm{\Omega }+9\mathrm{\Omega }+6\mathrm{\Omega }+0.5\mathrm{\Omega }+0.5\mathrm{\Omega }} \\ =0.167\mathrm{A} \end{gathered}$$

Now, the potential difference$V_{ad}$ is obtained as follows:

$$\begin{gathered} V_{ad}=8.0\mathrm{V}-\left(0.5\mathrm{\Omega }\right)\left(\frac{1}{6}\mathrm{A}\right)-\left(8\mathrm{\Omega }\right)\left(\frac{1}{6}\mathrm{A}\right) \\ =6.58\mathrm{V} \end{gathered}$$

Hence, the potential difference $V_{ad}$in the circuit is 6.58 V .

(b)

Consider the value of the internal resistance and the emf is:

$$\begin{gathered} r=0.5\mathrm{\Omega } \\ \epsilon =4.0\mathrm{V} \end{gathered}$$

Now, the potential difference $V_{bc}$is:

$V_{bc}=\epsilon -I\left(r\right)$

Substitute the values and solve.

$$\begin{gathered} V_{bc}=4.0\mathrm{V}-\left(0.167\mathrm{A}\right)\left(0.5\mathrm{\Omega }\right) \\ =4.08\mathrm{V} \end{gathered}$$

Hence, the terminal voltage of the $4.00\mathrm{V}$battery is $4.08\mathrm{V}$.

(c)

Given the emf of the battery is $\epsilon =4.0\mathrm{V}$

Determine $V_{ab}$battery with emf 10.30 V and internal resistance $r=0.5\mathrm{\Omega }$is inserted in the circuit at $d$.

So, in this case the direction of current is counter clockwise where current flows from negative terminal to positive terminal of 10.30 V battery. Therefore, the direction of 10.30 V battery will be the opposite of 8.0 V battery.

Now, the new current in the circuit is by Ohm’s law is as follows:

$$\begin{gathered} I=\frac{V}{R} \\ =\frac{10.30\mathrm{V}-8\hspace{0.17em}\mathrm{V}-4\mathrm{V}}{8\mathrm{\Omega }+9\mathrm{\Omega }+6\mathrm{\Omega }+0.5\mathrm{\Omega }+0.5\mathrm{\Omega }} \\ =0.257\mathrm{A} \end{gathered}$$

And, the potential difference $V_{bc}$is:

$$\begin{gathered} V_{ad}=\epsilon -I\left(r\right) \\ =4.0-0.257\left(0.5\right) \\ =3.87\mathrm{V} \end{gathered}$$

Hence, the difference of potential $V_{bc}$between the terminals of the 4.00 V battery is 3.87 V .