The circuit shown in Fig. E25.30 contains two batteries, each with an emf and an internal resist

step 1 We're given the circuit diagram on the left. In the book, instead of labeling these E1, R2, R3,

The circuit shown in Fig. E25.30 contains two batteries,...

Key Concepts

Kirchhoff's Loop Rule

This principle states that the sum of the potential differences (voltage gains and drops) around any closed circuit loop must equal zero. It is fundamental in analyzing circuits with multiple batteries and resistors as it allows the setup of equations that relate the emfs, internal resistances, and other voltage drops, leading to the calculation of unknown currents and voltages.

Ohm's Law

Ohm's Law defines the relationship between voltage, current, and resistance in an electrical circuit, stating that the voltage drop across a resistor is the product of the current through it and its resistance. This concept is essential for determining the voltage drops across circuit elements such as resistors and the internal resistances of batteries.

Battery Internal Resistance

Internal resistance is an inherent property of batteries that causes a voltage drop within the battery itself when current flows. This drop affects the terminal voltage, which is the voltage available to the external circuit, making it lower than the battery's emf. Understanding and including internal resistance is vital for accurately analyzing real-world circuits.

Terminal Voltage

Terminal voltage is the voltage measured across the terminals of a battery when it is connected to a circuit. It is determined by subtracting the voltage drop due to the battery’s internal resistance (current times internal resistance) from the emf. This concept is crucial for evaluating the effective voltage that drives current through the external components.

Potential Difference and Voltage Drops

The potential difference between two points in a circuit arises from the combination of voltage rises (such as those provided by batteries) and voltage drops (across resistors and internal resistances). Analyzing these potentials is necessary for understanding how energy is distributed throughout the circuit and for graphically representing the variations in electric potential along a path.

Transcript

00:01 We're given the circuit diagram on the left.

00:04 In the book, instead of labeling these e1, r2, r3, e2, r4, they put the actual numbers on the circuit, but i labeled them this way because it's a little bit easier to work with.

00:20 And the values are given above where the emf for battery 1 is 16 volts, the internal resistance associated with that battery is 1 .6 oms.

00:33 The resistor r2 has a value of 5 oms.

00:37 The internal resistance of the battery below r3 has 1 .4 oms, and the associated emf is 8 volts.

00:51 And the final resistor has a value of 9 oms.

00:55 In part a of this question, we're asked to find the current in the circuit, both its magnitude and its direction.

01:06 So there are two batteries here, which means that the current can be flowing in one of two ways.

01:16 It can either be flowing in the clockwise or to counterclockwise direction.

01:24 So if it's flowing into counterclockwise, it'll be flowing from b to a to c, which i forgot is a level here.

01:35 So it's just going around this way if it's counterclockwise.

01:38 And if it's clockwise, it's the other way around a to b to c.

01:42 So i'm going to make a guess, and i'm going to assume that the current is flowing counterclockwise.

01:50 And it actually doesn't matter which direction i pick because the sign of the current will tell me whether or not i pick the right direction or if it's the opposite direction.

02:03 And what i mean is when i solve for the current, say i assume that it's going counterclockwise.

02:10 If i get a positive value for the current when i solve for it, then that means this direction that i've picked is the correct one.

02:18 And if i get a negative value, then that means the direction that i picked, the current is actually flowing in the opposite direction.

02:26 So if i turned out to be negative, then it'll be flowing clockwise instead of counterclockwise.

02:33 So the way to solve for i is to remember that if you start at a certain point in the circuit and you come back and you make a full round trip so that you end where you started, then the total.

02:48 Potential difference the sum of all the potential differences that you encounter is zero um so let's go across uh we start here um at at the first battery and then we'll just gonna make our our way around in a counterclockwise direction so we encounter uh emf one and then um then there's r1 with the current i so there's that voltage drop there and then another voltage drop of r2i right as we go across here and then the next resistor we see is r3 so the voltage drop is r3i and now we're going we're going across you know we're crossing this other this other battery here but this is the the positive side and the negative side and the way that we determined a sign is when we go from negative to positive, that's a plus sign.

03:58 When we go from the higher potential end to lower potential, and over here, we're going in the opposite way.

04:05 We're going from the lower potential to higher potential in terms of the battery.

04:12 So we're crossing the battery in the opposite way.

04:16 So we have to put a minus sign to account for that.

04:19 And then we have a voltage drop here, minus for i and that's equal to zero so that's the total the sum of all the voltages or your potential differences in the circuit so we want to solve for i so let's actually collect all the terms with i and rewrite this equation so minus i and this here will just be the sum of all the resistance that's in the circuit.

05:04 And we can easily solve this equation for i.

05:08 So it's just the difference in the emf over the sum of the resistance.

05:22 And we actually have all the numbers.

05:25 So that's 16 volts minus 8 volts over r1, which is 1 .6 oms, and r2, which is 5, oms and r3 1 .4 oms and finally r4 which is 9 oms so we find that the current is 0 .47 amperors now we'll move on to part b okay so part b we're asked to uh so find the terminal voltage vab of the 16 -fold battery.

06:19 So let me quickly just give a sketch again of the circuit so we can have our points of reference.

06:39 So remember this is b, this is a, this is e1, this is r1, and yeah, that's all i need for this particular part of the problem.

06:54 So i'm not going to label in the rest but what we're going to do is we're start at bb and we're going to go over here to va so i'm at bb and then i pick up a contribution from the battery and then i have a voltage drop here of r1i and that should be equal to the a and then vab is the difference between va and vb.

07:44 But va is given by all of this stuff over here.

07:48 So vab is equal to vb plus e1 minus r1i and then vb.

08:04 So all of this is va...

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