A triangular array of resistors is shown in Fig. E26.5.

Question

(d) If the battery has an internal resistance of 3.00Ω, what current will the array draw if the battery is connected across bc?

Accepted Answer

Hey everyone. So today we're dealing with the problem about circuits. So we're being told that we have a circuit board below that is connected to three separate resistors. And we're going to have a 24 volt battery with an internal resistance of 4. homes to be connected across this series or this circuit. In two different scenarios when it's connected across RS. R. S. And across the point P. S. As such for being asked to find the current supplied by the battery in the circuit. So let's go ahead and look at scenario one. So for scenario one, if we connect this in, if we connected across Rs then that means that we have a voltage source, we'd have a voltage source as such, which makes the 13 OEMs and 11 ohm resistors in series while the nine ohm resistor becomes parallel to the entire circuit. The internal resistance Of the battery also is in series with the other two. So for chance one or for scenario one across R. S. The equivalent resistance of the other two resistors. It's the ones in series is equal to OEMs or is equal to the sum of the resistance in series so far. So it'll be our 11 plus R which are 11 and 13 arms respectively. Oops Plus 13. And I'll leave that there for now. For reasons you'll see in just a second for parallel series, we need to do one over are and we're doing equivalent resistance because now we're factoring in the two resistors in series as well as the one that's parallel to them now. So it'll be one over the nine own Resistor Plus one over The plus 13 homes equal to 1/9. OEMs plus 1/24. And when we simplify this we get 11/72. However, this is one over the equivalent resistance equal to alone over 72. Therefore the equivalent resistance becomes 72/11, Which simplifies to 6.55 homes. Now that we have the equivalent resistance of all the resistors in the series, we can go ahead and add the 6.55 homes to the four 20. OEMs from the from the voltage source because it's connected in series with everything else. Giving us a final answer of 10.75. So with the OEMs or with the resistance in the circuit now found. We can go ahead and use OEMs Law. We can go ahead and use OEMs Law which states that let's write this up here own slaw, which states that voltage is equal to I times R. Where I is current and are as resistance. So solving for I, the current we get I is equal to V by R. Which is equal to 10. M. S. Divided by a total resistance. Oops, my bad. This should be 24V, apologies, divided by the total resistance. We just found 10.75. Giving a final answer. Uh Oops giving us a final answer Of 2.23 Amps. 2.23 Amps. So that's scenario one. So let's write that up here for scenario one, the current supplied by the battery Will be 2.23 amps. Let's go ahead and erase everything here And get started on scenario two. So for scenario two, it's very similar. We use the same process. But now the position of the voltage source has changed instead of being across Rs is now across PS. Right, so it's across PS. Now the voltage source vaulted towards goes across here, making the 11 ohm resistor parallel To the 13 and nine resistors that are connected now in series with the voltage source. So we can use a similar process as we did earlier for two. This becomes the uh serious resistance So far becomes nine OEMs plus 13 OEMs. Excuse me. And the total resistance, The equivalent resistance becomes 1/11 OEMs Plus 1/22 homes. Which gives us an answer of 3/22. However, to find the equivalent resistance, we need to take the inverse Current resistance is equal to 22/3 or 7.33 ohms. Which we can then add 4.2 ohms. The internal resistance of the battery too because the battery is connected in series with the rest of the circuit, Which gives us an answer of 11. homes. Again using Arms Law. We know that i is equal to V over r The voltage is 24V. The resistance is 11.53 Os Giving a final current of 2.08 Amps. So for scenario two, the current applied across the circuit Is 2.08 Amps. So the answer choice that has both of these options, correct, is answer choice. D. Oops. I hope this helps. And I look forward to seeing you all in the next one.

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A triangular array of resistors is shown in Fig. E26.5.
(d) If the battery has an internal resistance of 3.00Ω, what current will the array draw if the battery is connected across bc?

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