Chapter 4: Q31E (page 843)

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).

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Step by step solution

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,

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

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

Take voltage at point a and b as $V_{a}$ and $V_{b}$ respectively.

$V_{a} = V_{b} + 16.0 V - l (1.6 Ω) So, V_{a} - V_{b} = V_{ab} = - 16.0 V + (1.41 A) (1.6 Ω) = 13.7 V . Thus, the voltage V_{ab} is - 13.7 V .$

Repeat similar calculation as part (b),

$V_{c} = - V_{a} + 16.0 V - l (1.6 Ω) - l (9.0 Ω) S o, V_{a c} = - 16.0 V + 15.0 V = - 1.0 V Thus, the voltage Vac is - 1.0 V .$

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

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