Batteries and Internal resistance

Click For Summary
SUMMARY

The discussion focuses on calculating the internal resistance of a 6.0 V battery, which is determined to be 0.556 ohms, and the external resistor value of 32.78 ohms when a current of 0.18 A is measured. The participants seek assistance in calculating the energy dissipated inside the battery over 4.4 minutes and the resulting current when a second identical battery is added in series with a 29-ohm resistor. Key concepts discussed include the use of Kirchhoff's voltage law to analyze the circuit with two batteries in series.

PREREQUISITES
  • Understanding of Ohm's Law
  • Familiarity with Kirchhoff's Voltage Law
  • Basic circuit analysis techniques
  • Knowledge of energy calculations in electrical circuits
NEXT STEPS
  • Learn how to calculate energy dissipated in resistors using the formula E = P * t
  • Study Kirchhoff's Voltage Law in detail to analyze series circuits
  • Explore the concept of internal resistance in batteries and its impact on circuit performance
  • Practice circuit analysis problems involving multiple batteries and resistors
USEFUL FOR

Students studying electrical engineering, physics enthusiasts, and anyone looking to deepen their understanding of circuit analysis and battery performance.

sabak22
Messages
23
Reaction score
0
A 6.0 V battery has an internal resistance r. The measured voltage (emf of the battery) is 6.0 V. When connected to a resistor R the terminal voltage is 5.9 V and the current is 0.18 A.
I managed to get a) What is the internal resistance r of the battery?
0.556 ohm

as well as b) What is the value of the external resistor R?
32.78 ohm

but i can't figure out what to do when they ask me
c) What is the energy dissipated inside the battery in 4.4 minutes?

and d)When a second identical battery is added in series and the external resistor is R = 29 Ohms what is the resulting current?

Please help me. I have been trying this for an hour!
 
Physics news on Phys.org
c) Power dissipated inside the battery would be

I^2r

where r is the internal resistance. once you get power, find the energy consumed inside the battery...

for part d) , set up Kirchhoff's equation again...
 
Thank you very much isaacNewton. But I am not quite sure how to solve the d part. Could you please expand a little bit? I am having trouble setting up the second identical battery in the circut.
 
have you learned how to use Kirchhoff's rule ?
 
umm all we've learned this week was that Iin=Iout. Is that supposed to help me? I don't know what to with it.
 
you should get some understanding of the voltage law to solve the problem
 
  • #10
just a quick question. will the two internal resistances be the same now that another identical battery has beenhadded? If yes then will the r= 0.556 like it was in the first case?
 
  • #11
yes, the internal resistance is the same... its the property of the battery...
but now you have two similar batteries in series...

Edit: which book your teacher is following ?
 
  • #12
so what I am getting is something like this :

6-0.556-32.78i1-6-0,556-29i2 = 0

Does this make any sense? if not can you please boost me with a little help.
 
  • #13
see, the batteries and the external resistor are in SERIES, so there is only one current throughout.
 
  • #14
draw a circuit to help you
 
  • #15
Moderator's note: thread moved to Homework & Coursework Questions.
 

Similar threads

  • · Replies 42 ·
2
Replies
42
Views
5K
  • · Replies 12 ·
Replies
12
Views
6K
  • · Replies 105 ·
4
Replies
105
Views
12K
Replies
15
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 11 ·
Replies
11
Views
6K