Solved example: Finding current & voltage in a circuit


– [Instructor] We have three
resistors connected as shown across a 50-volt supply. The question which I’ve not
written down to save space is to find the voltage
across each resistor and to find the current
through each resistor. Now before we start solving this, let’s quickly go through a common mistake that I would do while
solving problems like this. What I would do is apply Ohm’s law to each resistor directly. So remember Ohm’s law? Ohm’s law says V equals I times R. And what I’m thinking over here or what I used to think over here is I already know the voltage is 50. So then, for two ohm resistor
to calculate the current here, I would substitute R as two, V
is 50, calculate the current. Then for 40 Ohm resistor,
I would put V is 50, that’s already given, R is 40. Calculate the current,
same thing over here. And we are done. We now know current through each resistor. But do you understand, that’s wrong. Why is that wrong? That’s because when we apply Ohm’s law, V, which is the voltage, is
the potential difference across that resistor. For example, if I’m using
this for two ohm resistor, then I need to know what’s
the potential difference across two ohms. But I don’t know what’s
the potential difference across two ohms, 50 volts
is the potential difference across these two points. This point has the same
voltage as this point and this point as the
same voltage as this point which means, I know the
potential difference across this and this point. I don’t know the potential
difference across two ohms. So I can’t apply it for two ohms. I don’t know the potential
difference across ten ohms. I can’t apply it for ten. R, I don’t know even here. And that’s why I can’t
directly solve the problem. If you substitute V as
50 for each resistor, we are implying that 50 volts
is the potential difference across each resistor
which is clearly wrong. And that’s why we can’t do it that way. So, what’s the correct way to do this? , The correct way to do this,
is since I know the voltage across these two points,
I need to first, calculate what is the equivalent
resistance of these three. I need to replace these three resistors with one single resistor. And once I have done
that, then I can go ahead and apply Ohm’s law and calculate it. So I need to reduce this circuit. And we have seen how to
reduce circuits like this in a previous video, so
it’ll be a great idea to first pause and see if
you can try this yourself. All right, let’s do this. So, all we need to do
is identify resistors in series and in parallel. Let’s start with two and ten. Do you think they are in series? They look like they’re in
series, but are they in series? The answer is no. The vertical test whether the
two resistors are in series or not, is remember that they
need to have the same current flowing through them. That’s the definition of series. So, I would imagine a small
current flowing over here and see if that entire current flows here. If it does, they are in series. If not, they’re not in series. So, imagine some current flowing here. Now, as the current goes
forward, notice there’s a branch. Because of that, some
current might flow up and the rest of the
current will flow here. And as a result, the current here and here may not be the same. So they are not in series with each other. But if you look at these two resistors, they are in parallel. How do I check whether two
resistors are in parallel? They need to have the
same voltage across them. Let’s quickly check that. This point has the same
voltage as this point because there are no resistors in between. This point has the same
voltage as this point because there are no resistors in between. A wire would always have
the same voltage anywhere. We’re assuming the wires
don’t have any resistances. And when there is no resistance,
the potential difference is always zero within a
wire across any two points in a wire, so the voltage is the same. And so notice that this voltage, the potential difference here is the same as potential difference here. And therefore, they are in
parallel with each other. But for example, if there
was a resistor over here, then these two voltages, these two points won’t have the same voltage and then they wouldn’t be in parallel. But anyways, these are in parallel and so we can go ahead
and replace this resistor with an equivalent resistance. So how do we calculate equal
resistance in parallel? Well the formula for equal
in resistance in parallel is one over R equivalent
is going to be one over R1, which is going to be one over 44S, one over R1, plus one over R2, which is going to be one over 10. One over 10. So let’s solve this. We have a common denominator of 40. This will be one plus,
after multiply this by four to get 40, so multiply the
numerator also by four. That gives me five over 40. And remember, this is
one over R equivalent. So R equivalent would be,
let’s write that down, the reciprocal of this. 40 over five. And that is eight ohms. So what we have calculated
is that these two resistors connected in parallel can be replaced by a single resistor of eight ohm. And nothing will change. Everything in the circuit
will remain the same. The current in the
circuit and the voltage, everything will remain the same. So let’s go ahead and do that. So what we’ll do is I’ll keep the rest of the circuit as it is. So let’s draw the rest
of the circuit as it is, but replace this combination with a single resistor of eight ohms. There it is. And now, these two resistors
are in series with each other. How do we check whether
they are in series or not? How do we confirm? They need to have the same
current flowing through them. So let’s imagine a current flowing here. And notice all the current will flow here. There are no branches right now. And therefore, they are in series. And when resistors are in
series, the equivalent resistance is just the sum of the
individual resistances. And therefore, I’ll not write it down. The equal end over here will
just the sum of these two, eight plus two, so that will be 10 ohms. And so again, we can now
replace these two resistors with a single resistor of 10 ohms. And keep the rest of the circuit as it is, so let’s do that. Here it is. And we are done reduction
because we have reduced the circuit to a single resistor. And now I know the voltage
across these two points, which is the same as the
voltage across this point, now I know this voltage is 50 volts. So I know, let’s write that down. This point, the voltage
between these two points is 50 volts, I know that. This is 50 volts. And so, for this equal end resistance, I can now go and apply
Ohm’s law and calculate the current through this resistance. And that’s what we will do next. So let’s get rid of
this to make some space. And let’s apply Ohm’s law here. So we know V is 50. V is 50. That’s equal to I times R, R is 10. R is 10, so I is 50 divided by 10, that’s going to be five amperes. So I is five. So the current in this
circuit is going to be five, this is positive, this is negative, so the current flows from positive down to the negative terminal. And so that’s five amperes. So the current flowing to
this resistor is five amperes. But hold on, our original
question is to calculate the current through each
of these three resistors and the voltage across
these three resistors. But what we have done now
is calculate the current in this equivalent resistance. How do we get from here to there? Well now the trick is, we
go backwards from here. That’s why it’s important
to write down each step. So here’s what I mean. If we go back from here
to here, this 10 ohms splits as two and eight. And this splitting is a series splitting, that’s how I like to think about it. And remember, in series,
the current is the same. So whatever current is
flowing here, the same current must flow through this resistor
and this resistor as well. Because in series,
current remains the same. So the moment I know that the
current here is five amps, I also know that the current
here and the current here, of course, it must be the same current, that is also five amperes. And once I know the current,
the next thing I will do immediately, is to calculate the voltage across those resistors. By again, applying Ohm’s law. So, over here, notice, I
know the current is five, the resistance is two, V equals IR, so the voltage here must be 10 volts. Let’s use the same color. So this voltage across this
resistance must be 10 volts. And similarly, the voltage
across this resistance, IR, five times eight, must be 40 volts. And just to confirm,
notice, 10 and 40 adds up to give us a total of 50. Makes sense, because from here to here, the total voltage must be 50 volts. What’s the next step? Let’s go backwards. We’re already done with these two ohms. We already know this is five amps, and we know the voltage here is 10 volt. That part is already done. What to do here? Well now, this eight
ohms splits as 40 and 10 as a parallel combination. It’s a parallel split, as I
would like to think about it. And remember in parallel,
they have the same voltage. So whatever is the voltage here must be the same voltage over here. So immediately I know
the voltage across this must be 40 volts and the voltage
here must also be 40 volts. Oops, wrong color, let’s
use the same color. So the voltage here must also be 40 volts. It’s a little shabby, but
hopefully the color helps you identify or differentiate between them. And now that I know the
voltage, again apply Ohm’s law, this time to calculate the current. So that’s the whole game over here. If you know voltage, you
calculate the current. If you know the current,
you calculate the voltage. So, in this resistor, the
resistance is 10, voltage is 40. So I is V or R. So 40 divided by 10, that’s
going to be four amps. So current here is going to be four amps. And over here, 40 divided by
40 is going to be one amp. One amp. And again, just to check,
see notice that the five amp is getting split as one amp and four amp. Four plus one is five. So again, this conforms that
whatever we did is right. And we have now solved the
problem because we know all the current through each resistor and we also know the voltage
across each resistor. And so, to summarize, whenever
we have question like this where we have bunch of resistors connected in some combination across some voltage, then as we calculate the
current and the voltage across each one, first we’ll reduce it to a single resistance. Then we’ll calculate the
current through that resistance and the voltage across that resistance. And then, we’ll keep backtracking. When we go back, if the
resistors split as series, then we know the current must be the same. And then we know the current, next step would be to calculate the voltage. If we go back and we find this
split as parallel resistors, then the voltage is the same. Then we use Ohm’s law to
calculate the current. And that’s how you keep on backtracking regardless of how
complicated the circuit is, as long as you can reduce
it to a single resistor and you write down all
the steps in between, that’s important, otherwise,
it becomes a little bit difficult to do this. As long as you have written all the steps as in you’ve drawn all the
subcircuits in between, we can always go back and keep doing this. Calculate the voltage and
calculate the current.

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