Easy circuit question but i get the wrong answer for PD

  • Thread starter Thread starter mr_coffee
  • Start date Start date
  • Tags Tags
    Circuit Pd
Click For Summary

Homework Help Overview

The discussion revolves around a circuit problem involving a car battery with a specified emf and internal resistance, focusing on the potential difference across its terminals during charging and discharging conditions.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the potential difference using the formula Vb - Va = E - Ir, leading to a result of 9V, which they believe to be incorrect. Other participants provide conflicting responses regarding the correct potential difference.

Discussion Status

Participants are exploring different interpretations of the problem, particularly regarding the effects of charging versus discharging on the potential difference. Some guidance has been offered regarding the direction of current and potential drops, but no consensus has been reached.

Contextual Notes

There is a mention of the battery being charged with a specific current and the implications of internal resistance on the potential difference, which may not have been fully considered by all participants. The discussion reflects uncertainty about the application of the formula in different scenarios.

mr_coffee
Messages
1,613
Reaction score
1
Hello everyone! I am stuck on this simple problem...
A car battery with a 12 V emf and an internal resistance of 0.050 is being charged with a current of 60 A.

(a) What is the potential difference V across its terminals?
wrong check mark V
(b) What is the rate Pr of energy dissipation inside the battery?
W
(c) At what rate is electrical energy being converted to chemical energy?
W
(d) When the battery is used to supply 60 A to the starter motor, what is V?
V
(e) What is Pr in this case?
W

I found the V = 9V;
I drew the simple circuit then Used
Vb-Va = E-Ir;
V = 12v -60*.050 = 9V, but it was wrong, any ideas? thanks
 
Physics news on Phys.org
pd = 12V

correct
 
pd = 12V

correct me if i am wrong
 
nope, its wrong it said
 
mr_coffee said:
A car battery with a 12 V emf and an internal resistance of 0.050 is being charged with a current of 60 A.
(a) What is the potential difference V across its terminals?
I found the V = 9V;
I drew the simple circuit then Used
Vb-Va = E-Ir;
V = 12v -60*.050 = 9V, but it was wrong, any ideas?
The car battery can be imagined as an ideal voltage source of emf 12 V and the internal resitance r=0.050 ohm connected in series. See the picture. A and B are the terminals of the battery, and X is the virtual junction of the virtual ideal source with the internal resistance.
The carging current flows into the battery to supply positive charges onto the positive electrode. It causes a potential drop of 0.05 * 60 = 3 V across the internal resistance. The potential drops along the direction of the current. So B is more positive than X. X is more positive than A. So the net potential difference across the terminals A, B of the real battery is 12+3=15 V during the process of charging.
Your formula is true when the battery supplies current to the starter motor. The direction of this current is reverse with respect to the charging current and the potential drops 3 V from X to B inside the battery. The terminal voltage is 9 V during discharge.
ehild
 
Last edited:

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 34 ·
2
Replies
34
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 20 ·
Replies
20
Views
5K