The current through the resistor in Fig. 19–77 is 3.10 mA.
Chapter 19, Problem 78GP(choose chapter or problem)
The current through the \(4.0-k \Omega\) resistor in Fig. 19–77 is 3.10 mA. What is the terminal voltage \(V_{b a}\) of the “unknown”battery? (There are two answers. Why?)
\(4.0-k \Omega\)
\(V_{b a}\)
\(3.2 k \Omega\)
\(1.0 \Omega\)
\(8.0 \Omega\)
\(12.0 \mathrm{~V}\)
Equation transcription:
Text transcription:
4.0-k Omega
V{b a}
3.2 k Omega
1.0 Omega
8.0 Omega
12.0{~V}
Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.
Becoming a subscriber
Or look for another answer