In the circuit shown, the ammeter reads two amperes when switch K is open. Neglecting the internal resistance of the battery, find the reading when switch K is closed.
Taking a look at the circuit, we see that it has voltage supplied by a battery. This battery drives current through the circuit, which moves in an anticlockwise direction. We also see two resistors of equal resistances called 𝑅 in two parallel branches of the circuit. The statement mentions the switch K, which we see as originally open. And finally, there is an ammeter, a device for measuring current, in part of the circuit.
We’re told that when switch K is open, like it is now, then the ammeter reads off a value of two amperes. That’s how much current is flowing through this part of the circuit. With this in mind, we want to solve for the ammeter reading when switch K is closed rather than open.
As we get started, let’s consider the path that current follows through the circuit when the switch is open as shown. We know that the anticlockwise travelling current will flow down until it reaches this junction in our circuit. If switch K were closed, then the current would have the option of travelling across what we’ll call the upper parallel branch of our circuit. But, of course, when that switch is open, there is no electrical pathway for the current to follow, so none of it goes that way.
Instead, 100 percent of the current travels down through what we’ll call the lower branch of the parallel circuit, passes through the resistor 𝑅, and then through the ammeter. As the current passes through, the ammeter responds with a reading, we’re told, of two amperes. The current then continues on until it reaches the negative terminal of the battery. When it comes to our battery, we don’t know how much potential difference it supplies. But let’s give whatever that amount is a name. Let’s call it capital 𝑉.
So then, we have a potential difference across our circuit which creates a flow of current which travels through a resistor 𝑅. This may bring to mind the mathematical relationship that connects these three quantities, voltage, resistance, and current. That relationship, of course, is Ohm’s law. This law says that the potential difference across a circuit is equal to the current in the circuit multiplied by its resistance.
We’re going to apply Ohm’s law to our circuit for the two cases, when the switch is open as well as when it’s closed. Since we’re already set up with the switch K being open. Let’s start there. When the switch K is open, the potential difference across our circuit, we know, is capital 𝑉. That’s the name we’ve given to it. Just as a side note, this is the potential difference in the circuit regardless of whether switch K is open or closed.
But anyway, according to Ohm’s law, this potential difference is equal to the current in the circuit, which we know has a value of two amperes, multiplied by the resistance in the circuit. Even though we have the particular value of circuit current in this circumstance, two amperes, let’s come up with a symbol to represent the current in the circuit. Since the switch is open, let’s call the current in the circuit 𝐼 sub o, o standing for open.
So, that’s the current in the circuit. Then, what’s the resistance when the switch is open? Looking around our circuit, we see the only resistor that the current travels through is the resistor capital 𝑅. This is the case because, remember, we’re neglecting the internal resistance of the battery. So, we now have an Ohm’s law equation for the condition when switch K is open. Our circuit has a potential difference 𝑉, a current 𝐼 sub o, which we know is equal to two amperes, and it has a resistance capital 𝑅.
Now that we figured this out, it’s time to close switch K. When we do this, we know that we’ve created another pathway for current to travel in the circuit. So, the way the current moves will change. Now with switch K is closed, when current moves around and reaches the junction between the two parallel branches, it has the option of taking the upper branch through switch K as well as the lower branch. The current that travels across the upper branch with switch K will travel across this resistor until it’s ready to meet back up with the current that traversed the lower branch.
And notice, it’s only the current that goes through the lower branch which passes through our ammeter A. The currents in the two branches join back up and then return to the negative terminal of the battery. Now that our switch is closed. Let’s write an Ohm’s law expression for this circuit.
Just like before, we’ll start out with the potential difference across this circuit, which is still capital 𝑉. That hasn’t changed with the closing of switch K. The next term we want to solve for is the current in the circuit. And at this point, we’ll have to be careful. When our switch K was open, the current in the circuit was the same everywhere. There were no point where it could divide off into different branches. But now that our switch is closed, the current does split off into two separate pathways at one point before rejoining back to its original strength.
When we write the current in this circuit in our Ohm’s law equation, let’s choose a point in the circuit where the current is its maximum strength. That is, we’ll pick a location before the current has been divided across the two parallel branches. Let’s call the current in this circuit right after the positive terminal of the battery 𝐼 sub c. And it’s that value we’ll use in our Ohm’s law equation.
And finally, we want to write in the equivalent resistance of this circuit. The only two resistances in the circuit are the resistances 𝑅 in the top and the bottom branch of our parallel section. We want to write a resistance value which is equal to the equivalent resistance of these two resistors arranged this way.
When we have two resistors arranged in parallel, there’s a special equation which tells us their overall, or their equivalent, resistance. In a case such as this, then the equivalent resistance of our two parallel resistors is equal to the product of their resistances divided by their sum.
In our case, it so happens that our two resistors have the same exact value, labelled capital 𝑅. So, that means that their equivalent resistance is equal to 𝑅 times 𝑅 divided by 𝑅 plus 𝑅. 𝑅 times 𝑅 is 𝑅 squared, and 𝑅 plus 𝑅 is equal to two 𝑅. And we see that one factor of 𝑅 in the numerator and denominator cancels out. All this to say, the equivalent resistance of our circuit when switch K is closed is equal to 𝑅 over two.
We’re now in a great position because we have completed Ohm’s law equations for our circuit when the switch is open as well as when it’s closed. Notice that there is a link between these two equations. They both have the same potential difference, what we’ve called capital 𝑉, which tells us that we’re able to equate the right sides of these two equations since they both equal 𝑉. This means that 𝐼 sub o, the current when the switch is open, times 𝑅 is equal to 𝐼 sub c, the current in the circuit when the switch is closed, times 𝑅 over two.
Seeing a common factor of 𝑅 on both sides, we can cancel out that term. And we see that 𝐼 sub o is equal to 𝐼 sub c divided by two. Or two times the total circuit current when the switch is open is equal to the total circuit current when the switch is closed. Recalling that 𝐼 sub o, the current in the circuit when the switch is open, is two amperes, we can substitute two amps in for 𝐼 sub o here. And multiplying two by two amps, we find a value of four amps. That’s 𝐼 sub c.
Now the question is, is this our answer? And here we have to be very careful. What we’ve solved for is the total current in the circuit when the switch K is closed. But that’s not exactly what we’re asked for in the question. The question asks for the ammeter reading when the switch K is closed. And as we saw, because when the switch is closed, the total circuit current divides up over the two parallel branches of the circuit. And only one of those branches passes through the ammeter, that must mean that the ammeter reading is not the same as 𝐼 sub c. It will be something less.
To solve for it, we need to figure out just how it is that the current 𝐼 sub c will divide up over the two branches of our circuit. The answer to that question comes from looking at the relative resistances of the two parallel branches.
Notice that both the top branch and the bottom branch of our circuit have the same resistance value, capital 𝑅. Now electrical current, just like a current of water moving through a stream or a river, will seek out the path of least resistance. That’s the path it prefers. So, as our current approaches this junction in the circuit, if one of the branches had less resistance than the other, then relatively more current would go to that lower-resistance branch.
But since, in our case, the branches have identical resistances, that means the current will evenly divide up across the two. 50 percent of it will go across our top branch, and 50 percent will continue on down to the bottom branch. From the current’s perspective, there is no distinction between the branches. They’re both equally difficult to navigate. This means that half of our total circuit current when the switch is closed, half of 𝐼 sub c, traverses our bottom branch and then passes on through the ammeter.
So, looking at our equation for 𝐼 sub c, to solve for the resistance that passes through the bottom branch of our circuit and, therefore, through the ammeter, we’ll divide both sides of this equation by two. This reflects the fact that our current evenly splits between the two parallel branches. The current 𝐼 sub c over two is the current that moves through the ammeter. And that’s equal to two amperes. And that is the answer to our question, the reading of the ammeter when switch K is closed. We see then that whether the switch K is open or closed, the ammeter reads the same value two amperes.