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### Course: Class 12 Physics (India) > Unit 3

Lesson 10: DC Circuit analysis# Current through resistor in parallel: Worked example

How to calculate the current through a resistor in parallel using equivalent resistance and Ohm's law.

## Want to join the conversation?

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Find the voltage across R2 in volts (with at least one digit after the decimal point).(1 vote) - PLEASE ANSWER!

How do you calculate ohms if the amperes is less than one? It's not going to be accurate. Is there any other way? Thanks.(1 vote) - When a 5Ω resistor and two resistors R1 and R2 of unknown value are used in a

circuit as shown below, if the values shown on ammeters A1 and A2 are respectively

500mA and 100mA, the resistance value of R2 should be (show all working)?(1 vote) - what if the 2 ohms is not there then how to find current through 6 ohms(0 votes)

## Video transcript

- [Instructor] So we have
an interesting circuit here. The goal of this video is to
figure out what is the current that flows through this six-ohm resistor? Pause this video, and see
if you can work through it. So the way that I am going to tackle it is first simplify the circuit. Take these two resistors in parallel, and think about what the
equivalent resistance would be. And we have seen that before. One over the equivalent resistance is going to be equal to one over 6.0 ohms plus one over 12.0 ohms. 1/6 is the same thing as two over 12. So 2/12 plus 1/12 is 3/12. And 3/12, you could view that as the same thing as one over 4.0 ohms. And so one over the equivalent resistance is equal to one over four ohms. Well, that means that
the equivalent resistance is four ohms. And so we can simplify our circuit now, where we replace these
two resistors in parallel with one resistor of the
equivalent resistance, and that is going to be
equal to a four-ohm resistor. Now the next thing we could
do is we could figure out what the current is through
this part of the circuit, which would be the same
thing as the current right over there. We could call that i sub one. And we can just use Ohm's law for that. We would have i sub one would
be equal to our voltage drop, which is 24 volts, 24 volts divided by the equivalent resistance of
these two resistors in series. And when you have resistors
in series, you just add 'em up to figure out the equivalent resistance. So this would be divided by 2.0 ohms plus four ohms, plus 4.0 ohms. 24 divided by two plus four, or 24 divided by six, is four. And since we're dealing
with two significant digits, it'll be 4.0. And we're talking about current, so this is 4.0 amps, or 4.0 amperes. Now how do we use that information to calculate this current right over here? We can call that i sub two. Now one way to think about it is what is going to be your voltage drop from this point to this point? If you know the voltage drop
from that point to that point and if you know that the
voltage drop from this point all the way down here is, then we can figure out what
the voltage drop from here to here is going to be. So let's do that. So the voltage drop across
this first resistor, remember, your change in voltage is just equal to your current
times your resistance. And so this is going to be, your current is going to be four amperes times your resistance is two ohms, times two ohms, which is going
to be equal to 8.0 volts. And so if the voltage
difference between that point and that point is 24 volts, which we know from this voltage source, but if we drop eight volts
as we go to this point, well, then the difference
between this point and this point or this point and this
point right over here, this has got to be a 16-volt drop. If our delta V across the
six-ohm resistor is equal to 16 volts, well, then we can use Ohm's
law again to figure out i two. I sub two is going to be
equal to our drop in voltage, so 16 volts, divided by this resistance, six ohms. And so what is this going to be equal to? 16 divided by six is 2 4/6 or 2 2/3, or 2.666666, and if we round
to two significant digits, you're going to have 2.7 amps. And so we just figured out
what we wanted to figure out. This right over here is
2.7 amps, or 2.7 amperes. But we can keep analyzing it. For fun, I encourage you to figure out what that current is now,
the current i sub three. And use the exact same technique. And one thing that you should
feel very comfortable of is that this current that is flowing through the first resistor,
that four-ampere current, that current gets split
between i two and i three. And so i two and i three should add up to the original four amps. So just thinking about it that way, if you do the same type
of analysis we just did, you should get 1.3 amps
for i three because 2.7 plus 1.3 is going to
be equal to four amps.