Question

The circuit shown in Fig. E25.33 contains two batteries, each with an emf and an internal resistance, and two resistors. Find (a) the current in the circuit (magnitude and direction) and (b) the terminal voltage V_abof the 16.0-V battery.

Fig. E25.33

Short answer

1. The current in the circuit is 0.471 A.
2. The terminal voltage $V_{ab}$of the 16.0 V battery is 15.2 V

Step-by-step solution

Concept of Kirchhoff’s loop rule.

Kirchhoff’s loop rule states that the addition of all the potential differences going around the closed loop is zero.

  (a) Determination of the current in the circuit.

Apply the Kirchhoff’s loop rule around the circuit in the direction of the flow of current,

$$\begin{gathered} +16.0V-8.0V-l(1.6\Omega +5\Omega +1.4\Omega +9.0\Omega )=0 \\ l=\frac{+16.0V-8.0V}{1.6\Omega +5\Omega +1.4\Omega +9.0\Omega } \\ =0.471A \\ \mathrm{The}\mathrm{calculated}\mathrm{value}\mathrm{of}\mathrm{I}\mathrm{is}\mathrm{positive}\mathrm{and}\mathrm{thus}\mathrm{the}\mathrm{current}\mathrm{flows}\mathrm{in}\mathrm{anti}-\mathrm{clockwise}\mathrm{direction}. \end{gathered}$$

The calculated value of I is positive and thus the current flows in anti-clockwise direction.

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

Take voltage at point a and b as respectively.

$$\begin{gathered} V_a=V_b+16.0V-l(1.6\Omega ) \\ So, \\ {V}_a-V_b=V_{ab} \\ =16.0V-(1.6\Omega )(0.47A) \\ =15.2V. \\ \mathrm{Thus},\mathrm{the}\mathrm{voltage}{V}_{ab}is15.2V. \end{gathered}$$