(II) Suppose two batteries, with unequal emfs of 2.00 V and 3....

Question

(II) Suppose two batteries, with unequal emfs of 2.00 V and 3.00 V, are connected as shown in Fig. 26–63. If each internal resistance is r = 0.350Ω and R = 4.00Ω, what is the voltage across the resistor R?

Accepted Answer

Welcome back. Everyone in this problem. A remote weather station is powered by two backup batteries with different voltage ratings to ensure reliability. The batteries have emfs of 2.2 and 3.5 volts and each has an internal resistance of 0.4 ohms. The batteries are connected in a circuit that powers a critical sensor through a resistor of 4.5 ohms, determine the voltage across the resistor. R that powers the sensor. Here. We have a diagram for a circuit and for our answer choices. A says it's 2.8 volts, B, 3.6 volts, C 5.8 volts and D 7.2 volts. Now, in order to figure out the voltage across our resistor here, OK, then if we look at our circuit, we know that our resistor R is in series with a parallel configuration of our batteries. OK. And we know then that the voltage across the resistor R is going to be equal to the total current, let's call it I multiplied by the value of our resistance R. OK. But the problem is here, even though we know R, we don't know what the current is that's flowing through the resistor. So let's think about all of the currents in our circuit. Now, if we let I be the current that's flowing through our resistor, then when it gets to this junction for our batteries, we could say that I one is the current that flows through the 2.2 volt battery. And I two flows through the 3.5 volt battery. If we can figure out the values of I one and I two, we can add them to get I and thus solve for the voltage across the resistor. VR to help us do that. We'll need to apply Kurt Ko's junction rule. Let's assume that both currents flow from the positive to the negative terminal of their respective batteries. Then at the junction where the currents meet then by Kirk's junction rule, let's just shorten it as K JL the total current I through the resistor is going to be the sum of I one and I two, we still can't get a value for I I one or I two yet. So we need to do some more and to do that, we can use keras loop rule. OK. Now let's call that KLL Kirko loop rule. Now we can apply Ko's loop rule to any of the loops in our circuit. Let's use it for the loop containing the 2.2 volt battery. In other words, we could think of it as this inner loop here. OK. Now, for this loop. Ok, then we know that 2.2 and we're gonna ignore our units for no, but 2.2 minus I want and know we're going, we're going in this direction, we're going counter clockwise, right? So 2.2 minus 0.4 multiplied by I one. Ok, minus 4.5 multiplied by I is going to be equal to zero. If we do our loop containing the 3.5 volt battery, we can think of that as our outer loop here. Look, it would look something like this. Ok. Then for that loop 3.5 minus 0.4 I two. Again, we're going in the same direction minus 4.5 I minus 4.5. I sorry. Yeah, is going to be equal to zero. No, we can simplify this even further because we know that I equals I one plus I two. So we can substitute I equals I one plus I two in both of our equations. OK? And now let's write this in red for 2.2 volt and in blue 43.5 volt. So now in red, this can be simplified to 2.2 minus 0.4 I one minus 4.5 multiplied by I one plus I two being equal to zero. OK, let me bring that over a bit to make some space. And if we simplify this even further, it will be 2.2 minus 4.9 I one minus 4.5 I two equal zero. Let's call this equation one. Now, if we do the same substitute I into it into the loop for the 3.5 volt battery, we're gonna get 3.5 minus 0.4 I two minus 4.5 multiplied by I one plus I two being equal to zero. Ok. And no, that's gonna simplify it to 3.5 minus 4.5 I one. OK? Minus 4.9 I two equals zero. Let's call that equation two. Now we can solve I one and I two by eliminating a variable. Now let's do that. OK? By multiplying our first equation by 4.9 and the second by 4.5 to eliminate I two. So eliminating I two, eliminating to solve for I one. OK? Then when we multiply equation on one by 4.9 we get it to be 10.78 minus 24.01 I one minus 22.05 I two equal to zero. OK. So that's one multiplied by 4.9. And if we do it to equation two, it's going to be 15, let me fix that here. 15.75. Oops 15.75 minus 20.25 I one. OK. Minus 22.05 I two being equal to zero. OK. So this is equation two multiplied by 4.5. Now, if we subtract que two from equation one, OK, then we should find that we get it to be equal to negative 4.97 minus 3.76. OK? I one equal to zero. And no, that means then that I one is going to be equal to negative 4.97 divided by 3.76 which is approximately equal to negative 1.32 pairs. So we have a value for I one. Now we can substitute that value for I one into equation one. OK, to solve for I two. OK. So no, any equation on one is going to become 2.2 minus 4.9 multiplied by negative 1.32 minus 4.5 multiplied by I two equals zero. So that's gonna be 2.2 plus 6.468 minus 4.5 I two equal to zero. Now, if we try to isolate here, that means we get 4.5 I two equal to 8.668. And thus, that tells us that I too then is going to be equal to 8.668 divided by 4.5 which is approximately equal to 1.926 pires. Now this is good because now we have I one and I two and with I one and I two, now we can go ahead and solve for I because remember I is the sum of those values. So now this means OK, that I is going to be equal to negative 1.32 pi plus 1.926 S which would be equal to 0.606 S. And if that's the value of I, that means we can solve for the voltage across resistor R because it will be equal to I 0.606 pires multiplied by 4.5 ohms. The value of R which is equal to 2.727 volts or approximately 2.8 volts to two significant figures. Therefore, the voltage across the resistor that powers the sensor is approximately 2.8 volts. If we look back at our answer choices that tells us a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

(II) Suppose two batteries, with unequal emfs of 2.00 V and 3.00 V, are connected as shown in Fig. 26–63. If each internal resistance is r = 0.350Ω and R = 4.00Ω, what is the voltage across the resistor R?

Key Concepts

Electromotive Force (emf)

Internal Resistance

Ohm's Law

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