Solution Found!
Solved: In the circuit shown in Fig. E25.30, the 16.0-V
Chapter 25, Problem 35E(choose chapter or problem)
In the circuit shown in Fig. E25.32, 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 and direction); (b) the terminal voltage \(V_{b a}\) of the 16.0-V battery; (c) the potential difference \(V_{a c}\) of point ɑ with respect to point c. (d) Graph the potential rises and drops in this circuit (see Fig.25.20)
Equation Transcription:
Text Transcription:
V_ba
V_ac
Questions & Answers
QUESTION:
In the circuit shown in Fig. E25.32, 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 and direction); (b) the terminal voltage \(V_{b a}\) of the 16.0-V battery; (c) the potential difference \(V_{a c}\) of point ɑ with respect to point c. (d) Graph the potential rises and drops in this circuit (see Fig.25.20)
Equation Transcription:
Text Transcription:
V_ba
V_ac
ANSWER:
Step 1 of 2
Here, we shall have to use Kirchhoff’s voltage law to calculate the current in the circuit and then we can calculate the rest of the required things.
Let be the current.
Given that the polarity of the 16.0 V battery is reversed.
(a) Therefore, applying Kirchhoff’s voltage law,
The current value is 1.41 V. It has a negative sign, meaning its direction is clockwise.