Easy circuit question but i get the wrong answer for PD

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    Circuit Pd
In summary, 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
  • #1
mr_coffee
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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
 
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  • #2
pd = 12V

correct
 
  • #3
pd = 12V

correct me if i am wrong
 
  • #4
nope, its wrong it said
 
  • #5
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:

1. What is PD in a circuit?

PD stands for potential difference, which is the difference in electrical potential between two points in a circuit. It is measured in volts (V) and is often referred to as voltage.

2. How do I calculate PD in a circuit?

To calculate PD, you need to use Ohm's Law, which states that PD is equal to the current (I) multiplied by the resistance (R) in the circuit. So, PD = I * R.

3. What causes the wrong answer for PD in a circuit?

The most common cause for getting the wrong answer for PD in a circuit is a miscalculation or error in using Ohm's Law. It is important to double check your calculations and make sure you are using the correct units.

4. Can a circuit have multiple PDs?

Yes, a circuit can have multiple PDs. This can happen when there are multiple voltage sources, such as batteries, in the circuit. In this case, you would need to calculate the individual PDs and then add them together to get the total PD in the circuit.

5. How can I troubleshoot for the wrong PD in a circuit?

If you are consistently getting the wrong PD in a circuit, there may be an issue with your equipment or connections. Check all connections and make sure they are secure. You may also want to use a multimeter to check the PD at different points in the circuit to pinpoint where the issue may be occurring.

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