A triangular array of resistors is shown in Fig. E26.5. What current will this array draw from a 35.0 V battery having negligible internal resistance if we connect it across ab?

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Hey everyone today, we're dealing with the problem about circuits. So we're being told that we have three resistors that are connected in a circuit board. Three of the resistors are connected there while connected with a low resistance wire as the 4th leg. No, we're being told that there is a 19 volt power supply that is going to be connected across two different points P. S. And across points P. Q. And we're being asked to determine the total current in the circuit in both of those scenarios. So let's go ahead and draw this out. We'll start with the first scenario when it's across PS. So drawing our battery in, we can see that once it's connected here, This effectively makes this resistor here. The 6.50m resistor parallel to the other two resistors that are connected now currently connected in series with the battery. So what do we do here? Well, we know we're going to need Homes Law because we're going to be dealing with voltage resistance and current. And OEMs Law tells us the relation that voltage is equal to current times resistance. Now, in scenario one, the first thing we should do is find out the equivalent resistance of the resistors in series which is simply the sum of said series resistors Which is homes plus nine homes. Now we can use the parallel resistor equations to find the total equivalent resistance. So it'll be one over 6.50ms plus 1/13 plus nine. OEMs Which simplifies to one over 6.5 plus 1/22. 1/22 Which gives us a final simplified answer of 57 over 286 homes. Now this is one over the equivalent resistance. So to find the equivalent resistance itself we need to take the inverse of the fraction we just got. So it'll be 286 homes divided by 57 which gives us an answer of 5.02 5. homes rearranging. OEMs law to solve for current tells us that i is equal to V over R we know the voltages 19 volts stated in the question stem And the resistance is what we just found is 5.02 and current across resistors in a series which is this equivalent resistance that we found now will always be the same throughout. So Excuse me, the final amperage here will be 3.78 amps. So for scenario one across ps the current is 3.78 amps. So with this in mind we can go ahead and use a similar process for the second scenario when it goes over or when the battery is applied across points P. Q. Let's do that right now just as before when it's connected over PQ it turns this 13 ohm resistor turns this 13 ohm resistor into a parallel resistor to the other two resistors that are connected in series. Now with the battery. So we use similar processes. First we find the series resistors equivalents Which is 6.5 plus nine and we can find the parallel equivalent to make everything or to get the equivalent resistance of the entire system. This will be 1/13 Plus one over 6.2 6.5 Plus nine, Simplifying. We get 1/13 Plus one over 15. which gives us an answer of 57 The seven over 403 homes. Again, we need to inverse this to get the proper equivalent resistance. It will be 403. OEMs 403 homes Over 57 which gives us an answer of 7.07. Again solving for I using OEMs law. Now we have our voltage which is 19V V divided by 7.07 homes giving a final Ambridge Of 2.69 Amps. So for scenario two for scenario two, The equivalent or the current through the system when it's connected over PQ is 2.69 amps. And this lines up with answer choice A I hope this helps. And I look forward to seeing you all in the next one

A triangular array of resistors is shown in Fig. E26.5. What current will this array draw from a 35.0 V battery having negligible internal resistance if we connect it across ab?

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Key Concepts

Ohm's Law

Series and Parallel Circuits

Kirchhoff's Circuit Laws