In the circuit shown, find the ammeter reading, neglecting the internal resistance of the two batteries.
Taking a look at this circuit, we see the two batteries, 𝑉 𝐵 one and 𝑉 𝐵 two, 12 volts and two volts, respectively. 𝑉 𝐵 one creates conventional current that travels in an anticlockwise direction while 𝑉 𝐵 two has a flipped polarity. It creates conventional current moving clockwise. Also in this circuit, we have three resistors of four ohms, two ohms, and three ohms. And in addition to that, we have three currents marked out for us, 𝐼 one up top, 𝐼 two through the middle branch, and 𝐼 three down at the bottom. Our goal is to figure out what the reading of the ammeter is. In other words, what’s the current passing through this part of the circuit.
As a first step, let’s figure out where currents 𝐼 one, 𝐼 two, and 𝐼 three are flowing. We can start with 𝐼 one up top. The arrow tells us that it’s moving in an anticlockwise direction. So in other words, 𝐼 one moves through the four-ohm resistor and then comes around this corner and down to this junction. Looking at currents 𝐼 two and 𝐼 three, we see that they’re both marked out as moving left to right. That tells us that the current 𝐼 one, when it reaches this branch point, splits up into two parts. One part becomes current 𝐼 two and moves along this middle branch. And the other part continues on down to the lower part of our circuit, through the ammeter, through the three-ohm resistor, through this battery, and then on up to meet back up with 𝐼 two.
After 𝐼 two has crossed the two-ohm resistor, it rejoins with current 𝐼 three. And combined, we know that these are equal to current 𝐼 one. We see now where in this circuit currents 𝐼 one, 𝐼 two, and 𝐼 three are flowing. And we see, furthermore, that it’s current 𝐼 three we want to solve for because that’s the one the ammeter reads. To do that, we’ll analyze this circuit from the perspective of one of Kirchhoff’s laws. We’ll do it from the perspective of his voltage law. This law tells us that if we consider any closed loop in a circuit, then if we add up, that’s what this summation symbol means, if we add all the sources of EMF in that closed loop, then that sum must itself be equal to the sum of all the potential differences across that same loop.
In our circuit, the two batteries provide the EMF. And the current moving through the resistors provides the change in potential. As we mentioned, Kirchhoff’s voltage law is specifically for application in closed current loops. Looking at our particular circuit, we see two such loops we can identify. The one is in the top section of our circuit. And the other is in the bottom section.
We can successfully apply Kirchhoff’s voltage law to each one of these loops independently. To do that, we’ll need to make a decision about which direction we’ll pass through each of these two loops, whether it’ll be anticlockwise like this or clockwise like this. The direction we choose won’t affect the answer we get in the end. Either one will work. That being said, why don’t we keep these directions we’ve established, anticlockwise in our first loop, which we’ll call loop one, and clockwise in our second loop. We’ll call that loop two.
So beginning with loop one, let’s move all the way around this loop, counting up all the EMF sources as well as all the potential differences. It doesn’t matter where in the loop we start. But just to start somewhere, let’s begin at the EMF source, the battery 𝑉 sub 𝐵 one. This battery supplies 12 volts of potential. And looking around the rest of our loop, we see that this is the only EMF source. That means that 12 volts is the sum of all the EMFs according to Kirchhoff’s voltage law. This law says then that, for this loop, 12 volts is equal to the sum of potential differences across the rest of the loop.
Normally, we would include the potential difference across the battery itself due to its internal resistance. But recall that, in this case, we’re told to neglect that resistance. That means we’ll continue on around our loop, moving again in an anticlockwise direction. As we do, we arrive at the four-ohm resistor. And we see that current 𝐼 one moves through this resistor. The potential difference across it then is 𝐼 one times four ohms. We then continue on through our loop, our current becoming 𝐼 two. We then reach the two-ohm resistor. And we know that 𝐼 two passes through this. And that means we have another term in our summation of the potential differences, 𝐼 two times two ohms. Once we’re past that resistor, we continue on until we reach back to our battery. We’re finished loop one. And this is the Kirchhoff’s voltage law equation for that loop.
Notice two things. There are two unknowns in this equation, 𝐼 one and 𝐼 two, and also that 𝐼 three, the current that we really want to solve for, isn’t here. So let’s continue on. And now, we’ll make an equation for loop two. As we said, in loop two, we’ll be moving in a clockwise direction. And once again, we can start at our EMF source, 𝑉 𝐵 two. That EMF is the only EMF source in loop two. So that means that two volts is equal to the sum of potential differences across the rest of the loop. Following along in this direction, we encounter the three-ohm resistor, which current 𝐼 three moves through.
But notice, the direction we’re moving through this loop, clockwise, is opposite the direction that 𝐼 three moves through this resistor. That’s not a problem though. All it means is that we put a minus sign in front of this term on the right-hand side of our expression. Then, as we continue on through our loop, circling through the middle branch of our circuit, our current becomes 𝐼 two. And we once again encounter the two-ohm resistor. This time though, we’re moving in the same direction as the current as it moves through the resistor. This means that we add this potential difference, 𝐼 two times two ohms. Having crossed through that resistor, we then continue on until we get back to the beginning of our loop, the battery supply 𝑉 𝐵 two.
We now have two equations, one for each loop in our circuit. And these equations are independent of one another. In addition to these two, there’s a third independent equation we can write. Notice that current 𝐼 one divides up into currents 𝐼 two and 𝐼 three. And that when currents 𝐼 three and 𝐼 two join back up, they form current 𝐼 one. This tells us that current 𝐼 one is equal to the sum of currents 𝐼 two and 𝐼 three. And now, notice, we have three equations, one, two, three, and three unknowns, 𝐼 one, 𝐼 two, and 𝐼 three. We’ll use this system of equations to solve for current 𝐼 three. One way we can start doing this is taking equation two and subtracting it from equation one. If we do that, the 𝐼 two current term cancels out. And we find a result of 10 volts being equal to 𝐼 one times four ohms plus 𝐼 three times three ohms.
As a next step, we can substitute in 𝐼 two plus 𝐼 three for 𝐼 one. And when we do that, the right-hand side of our equation becomes 𝐼 two times four ohms plus 𝐼 three times seven ohms. Then, if we take equation two once more and multiply it by negative two on both sides, then we get this equation. Negative four volts is equal to negative two 𝐼 times four ohms plus 𝐼 three times six ohms. Adding these two equations together removes the 𝐼 two term. And we find a result of six volts being equal to 𝐼 three times 13 ohms. In other words, current 𝐼 three is equal to six volts divided by 13 ohms.
When we calculate this fraction, we find it’s 0.46 amperes. This is current 𝐼 three. And as we mentioned before, 𝐼 three is the current that moves through the ammeter. Therefore, this is also the ammeter reading.