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Internal resistance of battery

  1. Jun 12, 2008 #1
    1. The problem statement, all variables and given/known data

    a battery with unknown internal resistance and open circuit potential (emf) is connected toa 3 ohm external resistor. the terminal potential is found to be 60 volts. when the same battery is connected to a 48 ohm resistor, the terminal potential is 96 volts. what is the internal resistance of the battery?

    2. Relevant equations

    electric potential V = IR where I is current, R is resistance

    current I = epsilon/R where epsilon is emf

    3. The attempt at a solution

    I_1 = V_1/R_1 = 60/3 = 20 ampere

    I_2 = V_2/R_2 = 96/48 = 2 ampere

    now that is as far as i got, to find R internal, do i use a similar formula to R = V/I where V and I are the change in them?

    how does the emf come into play?
  2. jcsd
  3. Jun 12, 2008 #2
    First, your analysis about the current is wrong--you never took into account the internal resistance of the battery. In actuality, the current is I = V / (R_int + R), where R is the value of the resistor and R_int is the internal resistance of the battery. Anyways, this is the wrong approach.
    The key equation is the "voltage divider" equation: V = V_bat [R/(R + R_int)]. In this case, V is the voltage, or potential difference, between a point in the circuit just above the added resistor and the ground. Now, your given two cases but there are two unknowns in each case: (1) the battery's voltage and (2) the internal resistance. Are any of these unknowns constant? If so, what can we do with our two equations obtained from both cases?
    I noticed that you posted a lot of questions about circuits. If your willing to learn, and take the time, I recommend downloading a SPICE application--with it you can build circuits for most homework problems and obtain the desired quantity and check your work accordingly. Good luck!
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