The circuit shown in Fig. E25.30 contains two batteries, each with an emf and an internal resistance, and two resistors. Find (a) the current in the circuit (magnitude and direction); (b) the terminal voltage V_abof the 16.0-V battery; (c) the potential difference V_acof point awith respect to point c. (d) Using Fig. 25.20 as a model, graph the potential rises and drops in this circuit.

Fig. E25.30.

The cumulative sum of the change in potential all around the closed circuit is zero. There is a potential drop of IR when the current flows through theresistor and this drop is compensated when the current flows through an emf$\epsilon$ in the – to + terminal direction.

The current flows from + terminal to – terminal and as 16.0 V is the greater emf, the current will flow the counter clockwise direction.

Evaluate the sum of potentials and putit equal to zero,

$$\begin{gathered} +16.0V-8.0V-I\left(1.6\Omega +5\Omega +1.4\Omega +9.0\Omega \right)=0 \\ I=\frac{+16.0V-8.0V}{\left(1.6\Omega +5\Omega +1.4\Omega +9.0\Omega \right)} \\ =0.47A \end{gathered}$$

Thus, the current flow through the circuit is $0.47A$

Take voltage at point a and b as respectively.

$V_a=V_b+16.0V-I\left(1.6\Omega \right)$

So,

$$\begin{gathered} V_a-V_b=V_{ab} \\ =16.0V-\left(1.6\Omega \right)\left(0.47A\right) \\ =15.2V \end{gathered}$$

Thus, the voltage V_ab is$15.2V$

Repeat similar calculation as part (b),

$V_a=V_c+8.0V+l\left(1.4\Omega +5.0\Omega \right)$

So,

$$\begin{gathered} V_{ac}=\left(5.0\Omega \right)\left(0.47A\right)+\left(1.4\Omega \right)\left(0.47A\right)+8.0V \\ =11.0V \end{gathered}$$

Thus, the voltage V_ac is $11.0V$

Voltage drop between the points b and c

$$\begin{gathered} V_{cb}=\left(0.47A\right)\left(9.0\Omega \right) \\ =4.2V \end{gathered}$$

From the calculated voltages above, the potential at point ais 15.2 V above the potential at pointband the potential at point cis 11.0 V below the potential at point a,

Therefore, the potential of point cis

$15.2V-11.0V=4.2V$,

So, the point c is 4.2 V above the potential of point b.

The graph,