Video Transcript
The circuit shown contains three resistors. Find the difference between the current through the 4.3-ohm resistor and the 3.9-ohm resistor.
Looking at our circuit, we see that it consists of a cell providing a potential difference of 7.5 volts, as well as three resistors. We can see that the positive terminal of our cell faces to the right. And this means that conventional current in this circuit will point in a clockwise direction. When the current gets to this junction point, right here, some of it will branch off and travel towards the 4.3-ohm resistor. And some of the current will continue on traveling through the 3.9-ohm resistor. After that, the divided current will rejoin, pass on through the 2.7-ohm resistor, and move to the negative terminal of the cell.
We see then that in this circuit, the 4.3-ohm and the 3.9-ohm resistors are in parallel with one another. To make this more clear, we can redraw our circuit this way. Our question asks us to solve for the difference between the current through the 4.3-ohm and the 3.9-ohm resistors. As charge flows in the circuit in a clockwise direction, when it reaches this branch point, more charge will flow towards the branch that has the lower resistance. We can say then that overall the current in the 3.9-ohm resistor will be greater than that in the 4.3-ohm resistor.
In fact, putting two resistors in parallel, like we see here, is a common method for dividing up current. There’s a rule, known as the current divider rule, that tells us how current does divide up over two parallel branches. Given a total circuit current, 𝐼 sub t, and a resistance in the one branch of 𝑅 sub x and a resistance in the other of 𝑅 sub y, the current that travels through the branch with the resistance 𝑅 sub x, we’ll call this current 𝐼 sub x, is given by this expression. The current in this particular branch is equal to the total circuit current 𝐼 sub t multiplied by the resistance of the other parallel branch divided by the sum of the resistances of the two branches.
This equation helps confirm what we saw earlier that the lower the resistance in a parallel branch, the more current travels through that branch. In this case, if the resistance of the other parallel branch 𝑅 sub y is high, that will make much of the total charge in this circuit tend towards the other lower resistance branch. This will result in an increase in the current in this branch, what we’ve called 𝐼 sub x. The equation for the current in the other parallel branch is similar to the first one. It’s equal to the total current, 𝐼 sub t, multiplied by the resistance of the other parallel branch divided by the sum of the resistances of the two branches.
Note that in our scenario, we know the resistances of these two branches, but we don’t know the overall circuit current 𝐼 sub t. Before we can solve for the difference between the current in these two resistors, we’ll need to know that total current. In our circuit overall, there is a total potential difference, a total current, and a total equivalent resistance. These quantities are related through Ohm’s law. This law tells us that the total potential difference across our circuit equals the total current in the circuit multiplied by the equivalent circuit resistance. We’re given 𝑉, the total potential difference, as 7.5 volts.
To solve for 𝐼 then, we’ll need to know the total equivalent circuit resistance. We’ll call that capital 𝑅. Clearing some space to work at the top of our screen, let’s note that as we think about the total resistance 𝑅 in our circuit, we have to consider resistors that are in parallel, the 4.3- and 3.9-ohm resistors as well as a resistor that’s in series with these ones, the 2.7-ohm resistor. If we call the effective resistance of our two parallel resistors 𝑅 sub p, then the general equation for 𝑅 sub p, given exactly two resistors in parallel, we’ve called them 𝑅 one and 𝑅 two here, is given by this expression: 𝑅 one times 𝑅 two divided by the sum of 𝑅 one and 𝑅 two.
We can write then that the equivalent resistance of our two resistors in parallel, 𝑅 sub p, equals 4.3 ohms times 3.9 ohms divided by 4.3 ohms plus 3.9 ohms. We can substitute this expression in for 𝑅 sub p in our equation for the total circuit resistance capital 𝑅.
In this expression, the reason we add our equivalent parallel resistance to a resistor of 2.7 ohms like this is because given two resistors, we’ll call them 𝑅 one and 𝑅 two in series, the total resistance of those two together, 𝑅 sub s, equals their sum. In our equation for the total circuit resistance, we can think of this value as one individual resistor that is in series with the 2.7-ohm resistor. This is why we add these two resistances together. If we calculate this resistance value 𝑅, we get approximately 4.7451 ohms.
Returning now to our Ohm’s law expression, if we divide both sides of this equation by the overall resistance 𝑅, then that factor of 𝑅 cancels on the right. And we find that 𝑉 divided by 𝑅 is equal to 𝐼. Or switching the equation around, 𝐼 equals 𝑉 divided by 𝑅. We now know both the values for 𝑉, that’s 7.5 volts, and 𝑅, the total resistance in our circuit, of about 4.7451 ohms. The total current 𝐼 in our circuit then equals this potential difference divided by this resistance. This fraction is equal to approximately 1.58 amperes. So then, 1.58 amperes is the total current in our circuit. And knowing this, as well as the values of the resistors in parallel, we can calculate the current through the 3.9-ohm resistor and subtract from it the current through the 4.3-ohm resistor.
By the way, the reason we’re subtracting the current through the 4.3-ohm resistor from the current through the 3.9-ohm resistor is because we know that this first value, the current through the 3.9-ohm resistor, will be greater because this resistance is smaller than 4.3 ohms. This subtraction will yield a positive number.
So, first, let’s write an expression for the current through the 3.9-ohm resistor using the current divider rule above. It’s equal to the total circuit current, 1.58 amperes, multiplied by the resistance of the opposite parallel branch, 4.3 ohms, divided by the linear sum of the resistance of each branch. 3.9 ohms plus 4.3 ohms is equal to 8.2 ohms. Notice that the units of ohms in numerator and denominator cancel out. And this part of our expression simplifies to a number in amperes.
So this is the current through the 3.9-ohm resistor. We’re interested in the difference between this current and that through the 4.3-ohm resistor. The current through the 4.3-ohm resistor equals 1.58 amperes multiplied by the resistance of the opposite branch, 3.9 ohms, divided by the sum of the resistances of the two branches. Once again, 3.9 ohms plus 4.3 ohms is 8.2 ohms. And the units of ohms cancel from numerator and denominator. We now have a complete expression for the difference between our two currents. Notice that both terms in this expression are multiplied by 1.58 amperes and divided by 8.2. This means we can factor out this fraction, 1.58 amperes divided by 8.2. It’s multiplying 4.3 minus 3.9 or 0.4.
When we enter this expression on our calculator, then rounding our answer to three decimal places, we get 0.077 amperes. We note that this is not the current through either one of our parallel branches but rather the difference between the currents in these branches. That difference to three decimal places is 0.077 amperes.
