A 12 V car battery dies not so much because its voltage drops but...

Question

A 12 V car battery dies not so much because its voltage drops but because chemical reactions increase its internal resistance. A good battery connected with jumper cables can both start the engine and recharge the dead battery. Consider the automotive circuit of FIGURE P28.64.b. How much current is the dead battery alone able to drive through the starter motor?

Accepted Answer

Welcome back, everyone. We are making observations about the following given circuit and we are tasked with finding what is going to be the current flowing through the 0.750 resistor at the bottom of our circuit. Before getting started here, I do wish to acknowledge the multiple choice answers over on the left hand side of the screen, those are going to be the values in which we strive for. So without further ado let us begin. Well, what we are going to be using here are Kirchhoff's two rules for circuits, both the junction rule and the loop rule. Now, I wish to establish first the direction of our current flow per segment of our circuit. So what I'm gonna do is I'm gonna say going into the junction where I just drew the dot I'm gonna say that that is current one and coming out of the junction is current two flowing through the center and current three which will flow through the 0.75 0 resistor. So really what we want to look for is current three. Now, first and foremost, let me use your chops junction rule at the junction. I just drew. What we can say is that current one will be equal to current two plus current three. Now, what I will go ahead and also do is establish our first loop for Kirchhoff's loop rule. I'm going to look at the top half of our circuit and I'm going to deem that loop one. This is going to be Kirchhoff's loop rule for loop one. What we have going around, we have six, the voltage of the battery minus I one multiplied by 0.05 which is the voltage of the 0.050 resistor minus I two multiplied by 1.75. The voltage of the 1.750 resistor minus 3.8 which is the voltage of this second battery down here. And according to Kirch's loop rule, all of this will be equal to zero. Great. So now let me go ahead and establish a second loop here. And in fact, I'm going to be looking at the outer loop, the large loop of the entire circuit and I will deem that loop two. Let's establish an equation for Kirchhoff's loop rule or loop two. What we have is six minus I one multiplied by 0.05 minus I three, multiplied by 0.75 the voltage of our 0.750 resistor and all of this will be equal to zero. Let me go ahead and change colors here real quick as we can now start making substitutions and solving for I three. What I'm gonna do is I'm going to substitute the equation for Kirchhoff's junction rule into our two equations. For Kirchhoff's loop rules. What we get as we get the following six minus I squared plus or sorry, I two plus I three multiplied by 0.05 minus I two multiplied by 1.75 minus 3.8 is equal to zero. Simplifying this a little bit. What we get is 2.2 minus I two multiplied by 1.8 minus I three multiplied by 0.05 is equal to zero. Right now. Let me substitute our second equation more. Substitute K 12 junction, roll into our second equation. What we get is six minus I two plus I three multiplied by 0.05 minus I three multiplied by 0.75 is equal to zero. Simplifying this equation what we get is six minus I two multiplied by 0.05 minus I three multiplied by 0.8 is equal to zero. Sorry, rather 0.75 is equal to zero. All right. So then what I can do is I am going to stall for I two using our second equation, those tracking six from both sides and adding the I three term to both sides. What we get is I two is equal to six minus I three multiplied by 0.75. Divided by 0.05. Now I'm gonna substitute this value for I two all the way into this equation over here. And what do we get? We get 2.2 minus six minus I three multiplied by 0.75 divided by 0.05 multiplied by 1. minus I three multiplied by 0.05 is equal to zero. Let me scroll down here just a little bit to give us some more room. Now, what we wanna do is we want to solve for I three. So simplifying this down a little bit, we get 28.75 multiplied by I three is equal to 2, 13.8 dividing both sides by 28.75. What we get is that I three is equal to 7. amps corresponding to our final answer. Choice of B Thank you all so much for watching. I hope this video helped. We will see you all in the next one.

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