i got part a i need part b. A.The car lead-acid battery has generally 12 V outpu

Question

i got part a i need part b. A.The car lead-acid battery has generally 12 V output voltage, and its internal resistance is so small that can be neglected. When starting the engine, the current of 311 A is flowing from the battery to supply the car starter (an electrical motor that starts the engine engine). How much power does the car starter consume when starting the engine? 3732 W B.Now let's try using a pack of AA batteries (1.2 V output voltage and 0.14 Ohm internal resistance each) to replace the lead-acid battery. In order to mach the car lead-acid battery, first 10 such AA batteries need to be connected in series to form a 12V cell, while 50 such cells need to be connected in parallel (as shown in picture above; the total of 500 AA batteries are used). Find the power that is supplied to the same car starter using this pack of AA batteries. (Hint: for the second question, you need to find the total voltage and the internal resistance of the whole pack; you also need to find the resistance of the car starter ; this can be done using the data of the first part of this problem)

Explanation / Answer

This is a bit of a long question. I'll try to help as much as I can without giving the answer. In the first situation, since we are neglecting the input resistance, the voltage across the car starter is equal to the voltage source (voltage around loop must equal zero). The current is also the same current from conservation of charge. Therefore, Power is just V*I = 3732, which you already got. In preparation for part two, we can also calculate the resistance in the car starter using Ohm's Law, noting that V= IR so R = V/I. In the second part, each cell contains 10 AA batteries in series. In series, voltages and resistances add, and you can simplify each cell to a single ideal voltage source of 12 V plus 10*.14 Ohms(the internal resistance of each battery). Once you have the voltage and internal resistance of each cell, you can solve for the internal resistance of the 50 cells in parallel. Voltages do not add in parallel, but the resistors can be combined into a single equivalent resistance along with 12v source. To add resistors in parallel, you use 1/R_eq = 1/(10*.14) + 1/(10*.14) + ..... for all fifty cells. Therefore you have 1/R_eq = 50/(10*.14). You can then solve for the Equivalent Resistor. You now have a circuit with a single 12V voltage source, an equivalent resistor, and the car starter, which is just another resistor. You can use the voltage divider rule to solve for the voltage across the car starter: V_starter = 12V(Rstarter/(Rsource + R starter)) Solve for current using Ohms law I = V/R Then solve for Power using P=IV Or skip solving for I and just solve for power using P= V^2/R Enjoy!

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