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.
In this question, we’re given two
different circuits containing different series and parallel combinations of
resistors. And we want to find the ratio of
the total current produced by the circuit in diagram (a) to the one in diagram
(b). For both circuits, we’re given the
value of the potential difference provided by a cell and the resistances of all the
resistors. Since we need to use this
information to determine a value of current, we should use Ohm’s law.
Recall that Ohm’s law is typically
written as 𝑉 equals 𝐼 times 𝑅, where 𝑉 is potential difference, 𝐼 is current,
and 𝑅 is resistance. Here, we’re interested in
calculating current. So let’s rearrange this equation to
make 𝐼 the subject. To do this, we simply divide both
sides of the equation by 𝑅, which gives 𝐼 equals 𝑉 divided by 𝑅. Now, the diagrams show us the
potential difference across each circuit. But in order to use Ohm’s law for
an entire circuit, we need to find its total resistance first. We’ll do this by calculating the
equivalent resistance of the different resistor combinations for each circuit.
We’ll begin by considering the
circuit shown in diagram (a). To help us stay organized, let’s
label the 3.5-ohm resistor as 𝑅 sub one, the 2.5-ohm resistor as 𝑅 sub two, and
the 1.5-ohm resistor as 𝑅 sub three. We can see that resistors 𝑅 one
and 𝑅 two are connected in parallel. But 𝑅 three is not connected in
parallel because it is on the same branch of the circuit as the cell. So, we’ll first find the equivalent
resistance of the two resistors in parallel and then combine that with the remaining
resistor that’s connected in series. This will give us the total
resistance of the circuit.
Recall that for any number of
resistors connected in parallel, their equivalent resistance equals the reciprocal
of one over 𝑅 one plus one over 𝑅 two and so on. Here, we can replace resistors 𝑅
one and 𝑅 two with an equivalent resistor that we’ll call 𝑅 sub 𝑥, whose
resistance is given by the expression 𝑅 𝑥 equals the reciprocal of one over 𝑅 one
plus one over 𝑅 two. By substituting in the values of 𝑅
one and 𝑅 two, we find that 𝑅 𝑥 equals the reciprocal of one over 3.5 ohms plus
one over 2.5 ohms, which comes out to 1.4583 and so on ohms. This is the equivalent resistance
of the two resistors that are connected in parallel.
Now that the circuit is modeled as
two resistors, 𝑅 𝑥 and 𝑅 three, that are connected in series, we should recall
that for any number of resistors in series, their equivalent resistance is just the
sum of the individual resistances, 𝑅 one plus 𝑅 two and so on. So we can replace resistors 𝑅 𝑥
and 𝑅 three with a single equivalent resistor that we’ll call 𝑅 sub 𝐴, whose
resistance is given by the expression 𝑅 𝐴 equals 𝑅 𝑥 plus 𝑅 three. By substituting in the values of 𝑅
𝑥 and 𝑅 three, we find that 𝑅 𝐴 equals 1.4583 ohms plus 1.5 ohms, which equals
2.9583 ohms. Because the resistor 𝑅 𝐴 has a
resistance that’s equivalent to the combination of the three resistors shown in
diagram (a), this is the circuit’s total resistance.
Now that we’ve found the total
resistance of this first circuit, let’s turn our attention to the circuit shown in
diagram (b). Again, we’ll label the 3.5-ohm
resistor as 𝑅 one, the 2.5-ohm resistor as 𝑅 two, and the 1.5-ohm resistor as 𝑅
three. Here, we see that the resistors 𝑅
two and 𝑅 three are connected in series and that 𝑅 one is connected to both of
them in parallel. To find the total resistance of the
circuit, we’ll start by finding the equivalent resistance of the two resistors in
series and then combine that with the remaining resistor that’s connected in
parallel. So let’s replace resistors 𝑅 two
and 𝑅 three with an equivalent resistor that we’ll call 𝑅 sub 𝑦, whose resistance
equals 𝑅 two plus 𝑅 three. Substituting in the values of 𝑅
two and 𝑅 three, we find that 𝑅 𝑦 equals 2.5 ohms plus 1.5 ohms, or four
ohms.
Now that the circuit is modeled as
consisting of two resistors, 𝑅 𝑦 and 𝑅 one, connected in parallel, we can replace
them with a single equivalent resistor, 𝑅 sub 𝐵, whose resistance is given by 𝑅
𝐵 equals the reciprocal of one over 𝑅 𝑦 plus one over 𝑅 one. By substituting in the values of 𝑅
𝑦 and 𝑅 one, we find that 𝑅 𝐵 equals the reciprocal of one over four ohms plus
one over 3.5 ohms, which comes out to 1.8666 and so on ohms. This is the total resistance of the
circuit shown in diagram (b).
Now that we know the total
resistance of each circuit, we’re ready to find the ratio of the total currents of
the circuit in diagram (a) to the circuit in diagram (b). This ratio is written as 𝐼 sub 𝐴
divided by 𝐼 sub 𝐵. Before we start to calculate,
though, let’s clear some room on screen and use Ohm’s law to make a substitution for
each of these current terms.
Since current equals potential
difference divided by resistance, the ratio becomes 𝑉 𝐴 over 𝑅 𝐴 divided by 𝑉
𝐵 over 𝑅 𝐵, where 𝑉 𝐴 and 𝑉 𝐵 are the values of potential difference provided
by the circuit’s cells. Notice that the potential
difference provided by the cell in each circuit is the same, or, in other words, 𝑉
𝐴 equals 𝑉 𝐵. This means we can cancel out both
of the potential difference terms in the ratio, so it becomes one over 𝑅 𝐴 divided
by one over 𝑅 𝐵. To simplify this, recall that to
divide by a fraction, we can simply flip the denominator and multiply it
instead. So the ratio can be written as just
𝑅 𝐵 over 𝑅 𝐴.
Finally, we’re ready to substitute
in these values to reach our final answer. 𝑅 𝐵 over 𝑅 𝐴 equals 1.8666 ohms
over 2.9583 ohms. Notice that units of ohms cancel
out of the expression entirely, which makes sense because ratios are typically
unitless numbers. Now, plugging this into a
calculator, we get a result of about 0.6310. Remember that we were told to round
the answer to two decimal places, so this comes out to 0.63.
So, we’ve found that 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) is 0.63.
