Question Video: Determining Current in a Combination Circuit Physics • 9th Grade

Consider the circuit shown below. What is the reading on the ammeter?

06:04

Video Transcript

Consider the circuit shown below. What is the reading on the ammeter? (A) Two 𝐼 over three, (B) 𝐼 over two, (C) 𝐼, (D) 𝐼 over three.

To find the reading on the ammeter, we will need to calculate the current in that branch of the circuit. To do this, we will need to use Kirchhoff’s laws. We will begin by labeling all the components on the circuit as follows. We’ve labeled the resistors as 𝑅 one, 𝑅 two, 𝑅 three, and 𝑅 four, and we’ve also labeled currents 𝐼 one and 𝐼 two in addition to the current 𝐼.

Now, recall that Kirchhoff’s first law states that the sum of the currents into a junction or node in a circuit must be the same as the sum of the currents out of the junction. We can see that the current 𝐼 comes into the junction highlighted in blue and the currents 𝐼 one and 𝐼 two go out of this same junction.

So, by using Kirchhoff’s first law, we find that 𝐼 is equal to 𝐼 one plus 𝐼 two. The reading on the ammeter will equal the current 𝐼 one. We can make 𝐼 one the subject of the equation by subtracting 𝐼 two from both sides. This gives us that 𝐼 one is equal to 𝐼 minus 𝐼 two. We’ll label this as equation one. To find the values for these currents, we will also need Kirchhoff’s second law. Recall that Kirchhoff’s second law states that the sum of the potential differences across each of the components in a loop is equal to zero.

To begin with, we’ll consider loop one in this circuit. We’ll label the potential difference across resistor 𝑅 one as 𝑉 sub 𝑅 one. Similarly, we’ll label the potential difference across resistor 𝑅 two as 𝑉 sub 𝑅 two. Then, by applying Kirchhoff’s second law to loop one, we find that 𝑉 minus 𝑉 sub 𝑅 one minus 𝑉 sub 𝑅 two is equal to zero. Notice that the potential difference provided by the cell is positive, while the potential difference across each of the resistors is negative.

We can get expressions for 𝑉 sub 𝑅 one and 𝑉 sub 𝑅 two by using Ohm’s law. Recall that Ohm’s law can be written as 𝑉 is equal to 𝐼 times 𝑅, where 𝑉 is the potential difference, 𝐼 is the current, and 𝑅 is the resistance. The current in resistor 𝑅 one is the current 𝐼, so the potential difference across resistor 𝑅 one is 𝑉 sub 𝑅 one is equal to 𝐼 times 𝑅 one. The current in resistor 𝑅 two is the current 𝐼 one, so the potential difference across resistor 𝑅 two will be 𝑉 sub 𝑅 two is equal to 𝐼 one times 𝑅 two. Then, by substituting these equations into the equation we obtained from Kirchhoff’s second law for loop one, we have the equation 𝑉 minus 𝐼 times 𝑅 one minus 𝐼 one times 𝑅 two is equal to zero.

We are given the resistances of both resistors, with 𝑅 one is equal to 12 ohms and 𝑅 two is equal to nine ohms. Substituting these values into the equation, we get that 𝑉 minus 12𝐼 minus nine 𝐼 one is equal to zero. We’ll label this as equation two.

Now we can consider loop two in this circuit. We can label the potential difference across resistor 𝑅 three as 𝑉 sub 𝑅 three and the potential difference across resistor 𝑅 four as 𝑉 sub 𝑅 four. Applying Kirchhoff’s second law to loop two, we find that 𝑉 minus 𝑉 sub 𝑅 one minus 𝑉 sub 𝑅 three minus 𝑉 sub 𝑅 four is equal to zero. The current in resistor one is still the current 𝐼.

So Ohm’s law tells us that the potential difference across resistor 𝑅 one is 𝑉 sub 𝑅 one is equal to 𝐼 times 𝑅 one. Then, the current in resistor 𝑅 three is the current 𝐼 two. Ohm’s law tells us that the potential difference across resistor 𝑅 three will be 𝑉 sub 𝑅 three is equal to 𝐼 two times 𝑅 three. Similarly, the current in resistor four is also equal to 𝐼 two. So, Ohm’s law tells us that the potential difference across resistor 𝑅 four will be 𝑉 sub 𝑅 four is equal to 𝐼 two times 𝑅 four.

Substituting these expressions into the equation we obtained from Kirchhoff’s second law for loop two, we have the equation 𝑉 minus 𝐼 times 𝑅 one minus 𝐼 two times 𝑅 three minus 𝐼 two times 𝑅 four is equal to zero. We are given the values of these resistances. 𝑅 one is equal to 12 ohms, 𝑅 three is equal to seven ohms, and 𝑅 four is equal to 11 ohms. Substituting those values into this equation, we get that 𝑉 minus 12𝐼 minus seven 𝐼 two minus 11𝐼 two is equal to zero. We can then simplify this to get 𝑉 minus 12𝐼 minus 18𝐼 two is equal to zero. We’ll label this as equation three.

We can now compare equations two and three. We have two simultaneous equations. So by subtracting equation two from equation three, we get this equation here. We can simplify this by expanding out this bracket to get 𝑉 minus 12𝐼 minus 18𝐼 two minus 𝑉 plus 12𝐼 plus nine 𝐼 one is equal to zero. This then simplifies to negative 18𝐼 two plus nine 𝐼 one equals zero, or nine 𝐼 one is equal to 18𝐼 two. Dividing both sides by 18, we find that 𝐼 two is equal to 𝐼 one divided by two.

We can now substitute 𝐼 one over two in place of 𝐼 two in equation one, which we found using Kirchhoff’s first law, in order to find the value of 𝐼 one. This gives us that 𝐼 one is equal to 𝐼 minus 𝐼 one over two. Adding 𝐼 one over two to both sides of the equation, we get that three 𝐼 one over two is equal to 𝐼. Dividing both sides by three over two, we get 𝐼 one is equal to two 𝐼 over three.

We’ve already seen that 𝐼 one is the current that passes through the ammeter, and so the reading on the ammeter will be equal to this current 𝐼 one. Since we’ve found that 𝐼 one is equal to two 𝐼 over three, we know that the ammeter reading must be two 𝐼 over three. Comparing this answer with the available answer options, we find that it matches option (A). The reading on the ammeter is equal to two 𝐼 over three.