Question

In the circuit shown in Fig. E25.30, the 16.0-V battery is removed and reinserted with the opposite polarity, so that its negative terminal is now next to point a. Find (a) the current in the circuit (magnitude anddirection); (b) the terminal voltage V_baof the 16.0-V battery; (c) the potential difference V_acof point awith respect to point c. (d) Graph the potential rises and drops in this circuit (see Fig. 25.20).

Short answer

1. The current in the circuit is 1.41 A.
2. The terminal voltage $V_{ba}$of the 16.0 V battery is
3. The potential difference $V_{ac}$of point a with respect to point c is
4. The required graph is shown.

Step-by-step solution

  (a) Determination of the current in the circuit.

The new circuit is,

The current flows from + terminal to – terminal, the current will flow the clockwise direction.

Evaluate the sum of potentials and put it equal to zero,

$$\begin{gathered} +16.0V+8.0V-l(1.6\Omega +5\Omega +1.4\Omega +9.0\Omega )=0 \\ l=\frac{+16.0V+8.0V}{1.6\Omega +5\Omega +1.4\Omega +9.0\Omega } \\ =\frac{24.0V}{17.0\Omega } \\ =1.41A \end{gathered}$$

Thus, the current flow through the circuit is 1.41A. Here, both the batteries are driving the current on the clockwise direction.

(b) Determination of the terminal voltage Vab of the 16.0 V battery.

Take voltage at point a and b as $V_a$and $V_b$respectively.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{a}}={\mathrm{V}}_{\mathrm{b}}+16.0\mathrm{V}-\mathrm{l}(1.6\mathrm{\Omega }) \\ \mathrm{So}, \\ {\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{b}}={\mathrm{V}}_{\mathrm{ab}} \\ =-16.0\mathrm{V}+(1.41\mathrm{A})(1.6\mathrm{\Omega }) \\ =13.7\mathrm{V}. \\ \mathrm{Thus},\mathrm{the}\mathrm{voltage}{\mathrm{V}}_{\mathrm{ab}}\mathrm{is}-13.7\mathrm{V}. \end{gathered}$$

(c) Determination of the potential difference Vac of point a with respect to point c.

Repeat similar calculation as part (b),

$$\begin{gathered} V_c=-V_a+16.0V-l(1.6\Omega )-l(9.0\Omega ) \\ So, \\ {V}_{ac}=-16.0V+15.0V \\ =-1.0V \\ \mathrm{Thus},\mathrm{the}\mathrm{voltage}\mathrm{Vac}\mathrm{is}-1.0V. \end{gathered}$$

 Graph for the potential rises and drops in this circuit.

The graph is sketched by taking point a as zero potential.