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 ac?

Question

Accepted Answer

Hey everyone. So today we're dealing with the problem about circuits and resistors. So we're given this circuit where there are four resistors connected in series. We're being asked to find the total current in the circuit. When a 24 volt battery is connected across two points in two different scenarios. The first scenario says between points Q&R. Or across points Q and R. And the second scenario says across points P and R. So let's look at Q and R. First it's a little easier. So if we have a voltage source, a battery that is connected across Q and R. We have no internal resistance in the battery because it is ideal in this case. But since it is connected in Across the terminals across the two points, The circuit has changed a little bit. This 15 Ohm resistor has now become parallel To the other three resistors, the 13, 14 and eight Ohm resistors that are in series with each other. They are in series with each other and they are in serious with the battery. So to find current, we need to utilize OEMs Law of course, OEMs Law which tells us the relation between voltage current and resistance and getting started on our solution, we have the equivalent resistance of all the series resistors will be the sum of each of their resistance is so it'll be R one plus R two plus R three is equal to 13 plus 14 plus eight and those are all in homes. The coolant resistance because we're dealing with parallel systems as well. We have 1/15 homes Plus 1/13 plus 14 it was eight which simplifies to 1/15 homes plus 1/35 homes Which gives to over 21. However this is one over the equal resistance. So the equivalent resistance will be the inverse which gives us uh 21/ Or 10.5 homes plugging this into Homes Law. And solving for the current we get that current is equal to V by R. V by R. Which is simply 24V By 10.5 Ohms giving a current of 2. and pierce. So in scenario one There is a current of 2.29 amps. now to look at scenario to scenario two is a little more interesting because the uh voltage source, the battery is being connected across points Q. Or P. And R. And what this will look like will be something like this. We'll have oops we have the battery but it'll be something like this a little nicer. But what this does is it sort of separates the circuit and in this case We can do it either way. But let's say that 14 or the resisters oops the resistors for 14 homes and eight homes are in series with the battery while D 13 ohm. And the 15 ohm resistors are parallel. So what does this mean? Well the let's write this out for the ones in series, it will be R one plus R. Two which is simply uh 14 plus eight. However, we have two separate parallel systems. Right? So if we take the equivalent resistance of everything, We get 1/13 plus plus The equivalent resistance of 14 plus eight, which simplifies to 1/28 plus 1/22 Which gives us an answer one simplifying of 25 over inverse ng to get the equivalent resistance, we get x 25. It gives us 12.32 homes homes once again utilizing OEMs law, then get I is equal to V by R which is equal to 24V by 12.32, 12.32 giving a final Amperage of 1. Amps. So for scenario two, I is equal to 1.95 amps which lines up with answer choice, 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 ac?

Verified video answer for a similar problem:

Key Concepts

Ohm's Law

Series and Parallel Circuits

Kirchhoff's Circuit Laws