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Battery that expires in a circuit

  • Thread starter vipboon
  • Start date
  • #1
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Problem 5: Current, Energy and Power A battery of emf ε has internal resistance i R ,
and let us suppose that it can provide the emf to a total charge Q before it expires.
Suppose that it is connected by wires with negligible resistance to an external (load) with
resistance L R .
a) What is the current in the circuit?
b) What value of L R maximizes the current extracted from the battery, and how
much chemical energy is generated in the battery before it expires?
c) What value of L R maximizes the total power delivered to the load, and how much
energy is delivered to the load before it expires? How does this compare to the
energy generated in the battery before it expires?
d) What value for the resistance in the load L R would you need if you want to
deliver 90% of the chemical energy generated in the battery to the load? What
current should flow? How does the power delivered to the load now compare to
the maximum power output you found in part c)?




2. Homework Equations
Kirchoff's rules
V = IR
I = -dq/dt




3. The Attempt at a Solution
I'm assuming for part a, a simple use of Ohm's Laws would yield I = (Emf / (Ri + RL)
I don't know how to treat a battery that expires. Is there a set of equations for it? Or should I treat it was if it is a charged capacitor? Even if it is a charged capacitor I'm still a little clueless as to how to start this and what equations to use.
 

Answers and Replies

  • #2
lightgrav
Homework Helper
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read it more carefully ...
this battery has constant Voltage , constant internal resistance , constant load.

so , looking at your equation for I , what value of RL makes it biggest?

What kind of Energy does charge obtain from the chemical reactions?
 
  • #3
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Ah I see, that makes it more straight forward. Given I can only get smaller as R_L increases then wouldn't the largest value of I arise from R_L having a value of 0? That appears trivial though. Also, I'm assuming the energy would simply be Q*emf?
 
Last edited:
  • #4
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Update, I think I figured this one out, please let me know if I made a mistake:
b) R_L = 0
E = Qemf
c) Took the derivative of the power and set it equal to 0 to get R_L = r. This would mean it gets half of the energy produced by the battery.
d) In order to get 9/10 of the energy produced by the battery R_L = 9r. The rest is self explanatory.

Thanks a bunch
 

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