Video Transcript
The resistor in the circuit shown
is powered by two batteries in parallel that have terminal voltages of 2.5 volts
each. What is the potential drop across
the resistor?
So here in this circuit, we have
these two batteries, here’s one and here’s the other, and they’re arranged in
parallel. And they supply voltage to this
resistor here so that there’s a potential drop of 𝑉 volts across the resistor. Our task is to solve for this
value. And to do this, we can recall
Kirchhoff’s voltage law. In words, this law says that the
sum of emfs of voltage sources across a current loop equals the sum of potential
drops.
So this is saying that anytime we
have a voltage source such as these two batteries in our circuit, then if we
consider those sources in the context of a given current loop, the emfs supplied by
the sources are equal to the sum of potential drops experienced throughout the rest
of the loop due to circuit components such as resistors. Especially in our situation,
considering only those voltage sources and circuit components that are part of the
same current loop is very important.
If we forget this part about the
current loop, we might look at this circuit and say we have 2.5 volts here and 2.5
volts here. And both batteries seem oriented in
the same direction, so we expect these voltages to combine. And so we might say then that 𝑉 is
equal to the sum of these two voltages, 5.0 volts.
But if we constrain ourselves to
thinking in terms of current loops, as the voltage law tells us, then we’ll come up
with a different response. In this circuit, there are two
separate loops that contain both a battery as well as this resistor. Here’s one of those loops that
includes the lower of the two batteries. And then, here’s the second
loop. As we analyze this circuit, we’ll
consider what’s going on in these two different loops separately.
So first, let’s consider the
interior loop. We’ll call this loop one. Following Kirchhoff’s voltage law,
this says that the sum of emfs of voltage sources, which for this particular loop we
can see is 2.5 volts, is equal — Kirchhoff’s voltage law says — to the sum of
potential drops across this loop. As we look at loop one, we see that
there’s only one circuit component where voltage drop can occur. It’s over this resistor here. And we’re told that that voltage
drop is some amount we can call capital 𝑉.
So then applying Kirchhoff’s
voltage law to this first loop, we get 2.5 volts is equal to 𝑉. There are no other components in
this current loop. So our answer is simply that 𝑉,
the potential drop across the resistor, is 2.5 volts. But now, what if we consider the
other loop, we can call that loop two, instead. If we did that and applied
Kirchhoff’s voltage law, once again our sum of emfs of voltage sources around this
loop would be 2.5 volts.
And also just like before, the only
component where this potential can drop is the resistor. And therefore, that potential drop,
capital 𝑉, must be equal to 2.5 volts. So no matter which of the two loops
we consider, we end up with the same answer that the potential drop across the
resistor is 2.5 volts.
