Answer to Question #14421 in Algebra for john

Question #14421
A real battery is not an ideal independent voltage source. A voltage source is an appropriate idealization of the battery if the load on the battery is negligible. A better model for a battery is a voltage source in series with an ideal linear resistor whose resistance varies with temperature. Sometimes even better models are required - this Energizer technical bulletin gives more information. However, let's use the simple model of a linear resistor in series with an ideal independent voltage source, as in the figure.

It is suggested in section 1.5.1 of the textbook that to increase the current-capacity of a battery without increasing the voltage at the terminals we can connect batteries of the same voltage in parallel. Let's examine this using our model.
Let's assume that both component batteries have the same voltage V1=V2=1.5. The internal resistances of small batteries are about 0.2Ω, but they vary a bit. Let's assume that R1=0.25 and R2=0.32. What is the open-circuit voltage (in Volts) V of the combination?


Now, suppose we short-circuit the compound battery. (This is very dangerous. NEVER do this to a large battery, such as a lead-acid battery in a car, or to a lithium-ion battery from your laptop. You MAY live to regret it, but you may not.) What is the current (in Amperes) you should expect to go through the short circuit?


We can think of this combination as a bigger battery of the same voltage as the two component batteries. What is the equivalent resistance (in Ohms) of the compound battery? (Hint: you have the voltage with nothing connected and the current when shorted out.)


Now, suppose that the voltages of the two component batteries are not quite the same. For example, suppose that V2=1.6. Then when we hook the two batteries together current will flow and the higher voltage battery will charge the lower voltage one. What is the current (in Amperes) that will flow?

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