### Video Transcript

The figure shown represents part of an electrical circuit in which the potential difference between the points b and c is three volts. What is the potential difference between the points a and d?

Looking at the figure, we see points a, b, c, and d marked out along this segment of a circle. We also notice that the circuit elements between these points, which are all resistors, all have the same value, capital π
.

In order to solve for the potential difference between a and d, either end of this circuit segment, weβll start by solving for the total resistance between points b and c and then solve for the total resistance between a and d.

To calculate the total resistance between points b and c in this circuit, weβll want to add these two resistors together as theyβre arranged in parallel. We recall that the rule for adding resistors arranged in parallel tells us that one over the total resistance equals the sum of the inverses of each individual resistor. We can say then that one over the total resistance between points b and c equals one over π
plus one over π
. This right-hand side simplifies to two over π
. And when we cross-multiply, we find that π
sub bc equals π
over two. Thatβs the total resistance between the points b and c in our circuit.

Knowing this, we now recall Ohmβs law, which tells us that the potential difference across a circuit or a segment of a circuit equals the resistance of that segment multiplied by the current running through it.

Weβre told in the problem statement that the potential difference between points b and c, that is, the potential drop across these resistors, is three volts. According to Ohmβs law, we can write that π sub bc is equal to πΌ, the current running through point b and point c, whatever that is, times π
sub bc. Or in another way, we can say that three volts is equal to πΌ times π
over two.

Weβll store that relationship for later, and now weβll go on to calculating the total resistance between points a and d of this circuit. We know that the resistance between points a and b is simply π
. And weβve solved for the resistance between points b and c and found that to be π
over two.

To solve for the resistance between points c and d, weβll once again use our parallel resistor addition rule. One over π
sub cd equals one over π
plus one over π
plus one over π
, which equals three over π
. And when we cross-multiply, we see that π
sub cd is equal to π
over three.

Now that weβve calculated the equivalent resistances of these segments of our circuit, we can add them together in order to solve for the total resistance between points a and d. To do this, weβll use the series resistor addition rule, which says that the total resistance of resistors arranged in series is equal to their linear sum.

The total resistance, π
sub ad, of this circuit segment is equal to π
plus π
over two plus π
over three. We can create a common denominator of six in each of these three terms. So when we add them together, we find a total resistance of eleven sixth π
.

Weβll now use Ohmβs law one more time to solve for the potential difference from a to d. Itβs equal to the current in this circuit segment times π
sub ad. Comparing this relationship to our relationship solved earlier, we see that if we multiply the resistance term π
over two by three divided by three, then it can be expressed equivalently as three π
over six. Three π
over six times πΌ is equal to three volts, which means that 11π
over six times πΌ must be equal to 11 volts. Thatβs the potential difference from point a to point d in this circuit.