### Video Transcript

In the circuit shown, what is the ratio between the ammeter reading π΄ one and the ammeter reading π΄ two?

We see these two ammeters located in the circuit in this figure. And looking more closely at this circuit, we see that it consists of a power supply with no internal resistance and two resistors, π
and two π
, arranged in parallel.

We can begin solving for π΄ one to π΄ two by calculating how much current is running through this circuit. To figure that out, weβll use Ohmβs law, which says that the potential difference in a circuit is equal to the current in that circuit multiplied by the total resistance. And weβll also use the fact that when two resistors are arranged in parallel, their total equivalent resistance is equal to the product of their two values divided by their sum.

The total resistance in the circuit then, π
sub π, is equal to π
times two π
, all divided by three π
. A factor of π
cancels from this expression, and we find that the total resistance of this circuit is two-thirds times π
.

Ohmβs law then tells us that π sub π΅, the power supplied by the battery, is equal to πΌ sub π, the total current in the circuit, times two-thirds π
. In other words, πΌ sub π is equal to three π sub π΅ over two π
. Itβs helpful to have solved for the total current in the circuit because this is the reading read out by ammeter π΄ one. All the current that passes through the circuit passes through this ammeter. So we can now write that π΄ one over π΄ two is equal to three ππ΅ divided by two π
, all divided by π΄ two, the ammeter reading in the second ammeter.

To figure out how much current runs through this branch of the parallel circuit, weβre helped by realizing that all ππ΅ volts of potential difference from the battery must be dropped over each of these two branches. Using Ohmβs law once more, that means that ππ΅ is equal to the current in the two π
resistor branch of the parallel circuit multiplied by that resistance value. And this implies that that current, πΌ sub two π
, is equal to ππ΅ divided by two π
. This current value is the value read out by ammeter π΄ two, and therefore we can substitute it in for π΄ two in our equation.

When we consider this fraction, we see that the factor ππ΅ over two π
cancels from both numerator and denominator. This leaves us with three over one. Thatβs the ratio of ammeter reading π΄ one to ammeter reading π΄ two.