The topic of the problem is current division. In this problem, it's a classic problem for current division, it's the single node pair circuit. So we have a current source in this problem, and we have two parallel resistors. And what we're looking for is how that current i sub t from the current source is divided up between the resistor R1 path and the resistor R2 path. We know that the current i t from the source is going to split when it reaches that parallel combination of resistors. An in fact, the largest amount of that current will go through the smallest resistor. So the path with the least resistance gets the highest amount of the current, the largest amount of the current. So with this classic problem, what we're going to do is we're going to solve for the current I1 through resistor R1. Similarly, we could find the resistance the current I2 through resistor R2. So in order to do this, we need to perform Kirchhoff's current law on this circuit. And we first notice that these two nodes is the single node pair circuit. We have a node at the top and the node at the bottom. So we're going to start with adding the current into the top node. And if we do that, we have i of t minus i1 of t minus i2 of t is equal to 0. We can rewrite that equation, and we have i(t) equal to i1(t) + i2(t) like that. And we also know that there's a voltage drop across this single node pair circuit, V(t) from top to bottom. Positive top, negative on the bottom. And that V(t) is the same voltage drop across each of these elements, across the current source, across the resistor R1, and across the resistor R2. So we can take the currents i1 and i2 and we can write those in terms of the voltage drop V of t and the respective resistances. So i1(t) becomes V(t) divided by R1 and i2(t) becomes V(t) divided by R2. So we can then continue with the problem and rewrite the equation, pulling out our V(t). And then finding a common denominator for those fractions containing R1 and R2, we have R1 plus R2 divided by R1 times R2 as our common denominator. We can then rewrite that equation in terms of V(t). And so V(t) is going to be equal to i(t) times the product of the two resistances, divided by the sum of the two resistances. And so we end up with an equation that looks like that for i(t). We also know that, if we're looking for i1(t), i1(t) is going to be equal to V(t) divided by R1. And we have an equation for V(t). And V(t) is i(t) times R1 times R2 divided by R1 plus R2. So if we substitute that into our equation for V(t), we have i1(t) equal to i(t) times R1 times R2 divided by R1 plus R2, and all that's divided by R1. So ultimately, i1(t), simplifying that equation, becomes i(t) times R2 divided by R1 plus R2. And that's kind of our final statement of current division for the single node pair circuit. The current through the resistor R1 is equal to the current source times resistance R2 divided by the sum of R1 plus R2. Similarly, we could have written the equation for i2(t) through resistance R2. And it's going to be i(t) times R1 divided by R1 plus R2.
