# Question Video: Determining the Ratio of Total Current In Two Circuits Physics • 9th Grade

What is the ratio of the total current produced by the circuit shown in diagram (a) to the total current produced by the circuit shown in diagram (b)? Give your answer to two decimal places.

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### Video Transcript

What is the ratio of the total current produced by the circuit shown in diagram (a) to the total current produced by the circuit shown in diagram (b)? Give your answer to two decimal places.

Looking at these diagrams, we see that both of them show us circuits with three resistors of 15, 12, and 25 ohms. Both the circuits also have a cell that provides a potential difference of 36 volts. In fact, the only difference between the circuit shown in diagram (a) and that shown in diagram (b) has to do with the location of the cell in the circuit, where itβs placed relative to the resistors. To see the effect this placement has on these two circuits, letβs consider the path of conventional current in each case.

Conventionally, current always leaves the positive terminal of a cell. In the circuit in diagram (a) then, current travels in this direction, reaches this junction point, splits across these two branches of the circuit, joins back together, and then arrives at the negative terminal of the 36-volt cell. What we have then is all the moving charge in the circuit passing through this 25-ohm resistor, but then that moving charge divides up across these two parallel branches so that only some of it passes through the 15-ohm resistor and the rest passes through the 12-ohm resistor. Letβs compare this movement of charge to the movement of electrical charge in the circuit in diagram (b).

Here, conventionally, that charge also leaves the positive terminal of the cell, but then it quickly reaches a junction. The current splits up, and some of the moving charge passes through the 25-ohm resistor, while the rest of it goes to pass through the 12-ohm resistor. These two separated currents join back up here and then continue on all together to pass through the 15-ohm resistor and finally to the negative terminal of the cell. In the circuit in diagram (b) then, all of the circuit current passes through the 15-ohm resistor, but only some of that current passes through the 12-ohm resistor with the rest passing through the 25-ohm resistor.

We want to solve for the ratio of the total current produced in each of these circuits. Letβs call the total current in the circuit shown in diagram (a) πΌ sub π and the total current in the circuit shown in diagram (b) πΌ sub π. Clearing some space on screen, what we want to solve for is πΌ sub π divided by πΌ sub π, that is, their ratio. We can work towards solving for this ratio using Ohmβs law. This law tells us that the potential difference across a closed electrical circuit is equal to the total current in that circuit multiplied by the total circuit resistance. Note that if we divide both sides of this equation by that resistance π, that factor will cancel on the right. And we see that πΌ is equal to π divided by π.

We can apply this form of Ohmβs law to our two circuits. The total current in circuit (a) πΌ sub π is equal to the total potential difference across that circuit π sub π divided by the total resistance in the circuit π sub π. Likewise for the circuit shown in diagram (b), πΌ sub π equals π sub π divided by π sub π. Both of these potential differences, π sub π and π sub π, have the same value. Theyβre are 36 volts. Our next step will be to solve for the total resistances π sub π and π sub π in these two circuits.

Letβs begin with π sub π. And we can note that these two resistors, the 15-ohm and the 12-ohm resistors, are in parallel with one another. We see this because the total current πΌ sub π has been divided across these two parallel branches before it reaches these resistors. We want to find the effective resistance of these two resistors in parallel and then combine that with this 25-ohm resistor. At this point, letβs recall that when exactly two resistors β weβll call them π one and π two β are arranged in parallel, then if we were to combine these resistors to have one equivalent resistance π sub π, it will be equal to the product of the resistances, π one times π two, divided by the sum of the resistances, π one plus π two.

If we find the effective resistance of our 15- and 12-ohm resistors in parallel in our circuit shown in diagram (a), then that will be equal then to 15 ohms times 12 ohms divided by 15 ohms plus 12 ohms. This takes care of two of the resistors in the circuit, but we still have to combine this resistance with our 25-ohm resistor. We know that this resistor is arranged in series with the others. Given two resistors arranged in series with one another, their total resistance β weβve called it π sub π  β is equal to the sum of their individual resistances.

In the circuit in diagram (a) then, if we find the equivalent resistance of our two parallel resistors and then add that value to 25 ohms, we will then have an expression for the total resistance π sub π in this circuit. If we go ahead and calculate π sub π, we get a result of 31.6 repeating ohms.

Knowing this, letβs now follow a similar process for the circuit shown in diagram (b). Notice that in this circuit the two resistors that are arranged in parallel with one another are the 25-ohm and 12-ohm resistors. So then, by our rule for finding the equivalent resistance of two resistors in parallel, we have the equivalent resistance of these two resistors. And then since theyβre arranged in series with the 15-ohm resistor, we add that resistance to this fraction. And by doing so, now we have an expression for the total resistance in the circuit π sub π. If we calculate this resistance, we find a value of 23.108 with the 108 repeating ohms.

Now that we know both π sub π and π sub π, we have all the information we need to calculate this fraction of interest. Before we substitute in our known values for π sub π and π sub π though, notice that if we take this fraction and we multiply both the numerator and the denominator by π sub π divided by 36 volts, then in the denominator the factor of 36 volts cancels out, as does π sub π. And in the numerator, 36 volts cancels from this numerator and denominator. What we end up with is a ratio of resistances. Notice that this ratio is flipped, we could say, from the ratio of currents. We can see the reason for this from Ohmβs law. Current and resistance are inversely proportional.

We now substitute in our known values for π sub π and π sub π. Note by the way that the unit of ohms in our numerator will cancel out with the unit of ohms in our denominator. This fraction will be a pure number. And when we calculate it to two decimal places, we get a result of 0.73. This is the ratio of the total current in the circuit shown in diagram (a) to the total current in the circuit shown in diagram (b).