The circuit shown in Fig. E25.33
contains two batteries, each with an emf and an internal resistance, and two resistors. Find (b) the terminal voltage Vab of the 16.0-V battery

Question

The circuit shown in Fig. E25.33

Circuit diagram showing two batteries, resistors, and their values for Kirchhoff's Loop Rule.

contains two batteries, each with an emf and an internal resistance, and two resistors. Find (b) the terminal voltage Vab of the 16.0-V battery

Accepted Answer

Hey everyone in this problem, we're told that non ideal power supplies are used to power the circuit shown the E. M. F. Of the power supply is nine volts and four volts. And were asked to determine the terminal voltage. Okay. Of the nine volt power supply. Alright, so let's recall that our terminal voltage. Okay, we can write this as V A B is equal to epsilon minus I times little R. Okay, now epsilon is gonna be the E. M. F. Of the source. Okay, we're told that we're using that nine volt power supply. So we know the value of ε is going to be 9V. R is going to be the internal resistance of the source. We know the internal resistance of the source. We can look at this diagram and determine that. Okay, and i is the current through our source. Okay, well we don't know the current. Okay, so how can we find it? Well, let's go ahead and use Kirchoff Law to try to find the current. I Okay, so Kirchoff law tells us that the sum of the voltage across the circuit in this loop is going to be equal to zero. So let's start at this nine volt source And work counterclockwise. Okay, so this is gonna be our direction of travel. We're going to assume that this is the direction of the current, so at our power supply here, we're going from the negative to positive. So that's going to give us a positive 9V and now we're going in the direction of the current or the direction we assume the current is going in. Okay, so all of these resistors are going to be negative, so we're going to have negative and when we run across the resistor, okay, we want to write the voltage, recall that the voltage is going to be I times are okay. The current times the resistance according to law. So we're gonna have I here Times the resistance of 1.5 homes. Alright then we're gonna continue along our loop. We hit another resistor. 10 OEMs, same thing. Okay, we're going in the direction of the current. So this is going to be negative and it's going to be the current. I times a resistor which is 10 times carrying along our loop. Counterclockwise, we hit another resistor 5.5 OEMs. So again, I times 5.5 bombs, Another resistor of one Ohm, I times one. And finally we hit this second power supply of four volts. Okay, when we're going counterclockwise we're going from the positive to the negative. Okay, so we're gonna be that's gonna be a negative voltage. So we're going to have negative four volts and all of this is going to equal to zero according to Kirchoff law. Alright, so you'll see that now the unknowns in this problem, Mark. I okay, we can factor this because this i is the same. The current is gonna be the same across all of these resistors. Okay, so this is the same. I so we can factor it So 9V -4V. We're going to have five volts. Okay, moving everything else to the right hand side, we're going to have I Times 1.50ms lost 10 OEMs Plus 5.5. OEMs last one owned. This is gonna give us I times 18 times. And if we divide we get eyes gonna be 5/18, we have volts divided by OEMs which is going to give us amperes the unit of current. Okay, Alright, so now we've found the current and we can get back to determining the terminal voltage that we were asked to find. That's what recall terminal voltage V A B is equal to epsilon minus I times little r epsilon Again, this is the E M F of the source. We're using the nine volt power supply. Okay, so this is going to be nine volts. We're going to subtract the current. We just found a 5/18 amps Times the resistance of the internal resistance of the source. Okay, now our power supply here is 9V in rent and you'll notice that this resistance is also in red. Okay, so that is the internal resistance of that power supply. Okay, that corresponding resistance and so our internal resistance is going to be 1.5 OEMs. So now we have nine volts minus 5/18 amps times 1.5 M. S. And this is going to give us a value of 8.5833 repeated. Okay, this first term we have volts, then we have amps times owns times is going to give us a volt so minus volt. So we end up with a unit of volts. Okay, so we have 8.5833 repeated is going to be our terminal voltage. So if you look at our answer choices, we see That are terminal voltage is going to be CK 8.58V. Thanks everyone for watching. I hope this video helped see you in the next one.

The circuit shown in Fig. E25.33
contains two batteries, each with an emf and an internal resistance, and two resistors. Find (b) the terminal voltage Vab of the 16.0-V battery

Verified Solution

Key Concepts

Kirchhoff's Loop Rule

Terminal Voltage

Ohm's Law

The circuit shown in Fig. E25.33 contains two batteries, each with an emf and an internal resistance, and two resistors. Find (b) the terminal voltage Vab of the 16.0-V battery

Verified Solution

Key Concepts

Kirchhoff's Loop Rule

Terminal Voltage

Ohm's Law

The circuit shown in Fig. E25.33
contains two batteries, each with an emf and an internal resistance, and two resistors. Find (b) the terminal voltage Vab of the 16.0-V battery