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 bc?

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Hey everybody today, we're dealing with problems about circuits. So we're being told that we have three resistors that are connected to form a circle and we're being asked to find the current and ideal, 26 volt battery delivers to the circuit when connected across two points in two different scenarios. The first is across points Q. And R. And the second is across points Q. And S. So let's take a look at the first scenario first. So let's just say that this is our, oops, let's do that in blue. Let's just say that we have our battery here and it's connected to point Q. And connects to point our This is scenario one q. R. But what this means is we have effectively made this 18 ohm resistor parallel to the other two resistors that are still in series with the battery. Now that we have this battery connected here. So how do we go about doing this? Well, to find the current, we're going to need Homes Law, which gives us the relation between voltage, current and resistance as voltage is the product of current and resistance. So going ahead with this, we should first find the equivalent resistance of the resistors in series which is as we know the some of the set resistors in series, This will be 22. OEMs plus 13 OEMs. Now the total resistance. Now that we're factoring in the one in parallel as well be 1/18. OEMs, oops plus one over 22 plus 13 homes Simplifying this, we get 1/18 plus 1/ 1/35. Giving us a final answer. Once we simplify of 53 over 630. No we can uh see that this equation that we just use for the parallel resistors gives one over the equivalent resistance. So we invert the answer To get the actual resistance. So this will be 6:30 Divided by 53 Which will give us a equivalent resistance of 11.89 homes. So with this and we have the voltage as 26V. We can rearrange OEMs law and solve for the current. So this gives us that i the current is equal to V by R. And substituting in our values. That's 26V over 11. homes Giving us a final current of 2.19 Amps. Let's write that up here For Scenario one. The current Is 2.19 Amps. So now that we know how this is done, we can go ahead and repeat the same steps or very similar steps for scenario 21 is connected across QS. So let's erase all of this real quick. So when it's connected across Qs and I'll just draw it inside here for space. But let's connect across points. Q. And s now uh the 22 ohm resistor has become parallel to the 13 and 18 ohm resistors. So we can go ahead and do very similar things. First we find the series resistors equivalent value. R. one plus R. two. Just simply 13 plus 18. Now we can try to do the parallel resistor equation to get the equal and resistance of everything. This will be 22 ohms plus 1/13 plus 18. Simplifying. We get 1/22 22 plus 1/31. Giving us a final answer after simplification of 53 over 682. Now we can go ahead and in verse this invert this sorry to get our equivalent equivalent is equal to 682 over Which gives us a final answer of 12.87 homes substituting this into Homes Law which states that the current is equal to V by R Our voltage is 26 volts and our resistance is 12.870. M. S. As we just found out we have a final current value Of 2.02 Amps. So option choice to i Is equal to 2. amperes. And this lines up with answer choice. See 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 bc?

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Key Concepts

Ohm's Law

Series and Parallel Circuits

Kirchhoff's Voltage Law