Here's a simple circuit containing two resistors, R1 and R2 in series. We can use this circuit as a potential divider or a voltage divider. Let's write down what we know about this circuit already. It's a series circuit, so the current through the circuit is the same at all places. So the current through R1, I, is the same as the current through R2, I, and that's the same as the current that returns to the battery. And now the voltage. The voltage of the supply I'll call VS. The voltage across R1, I'll call V1 and V2 for the voltage across R2. We know for a series circuit that the voltages of the components add up to the voltage of the supply. So VS is equal to V1 plus V2. For any component, the voltage across that component is equal to the current through that component multiplied by its resistance. V equals IR. Let's use this to examine the voltages in these components. V1 is equal to the current through R1, which is I, multiplied by its resistance. And V2 equals the same current I multiplied by R2. With potential divider circuits, we're interested in the ratio of the voltages. V1 divided by V2. V1 is equal to I times R1. And V2 is I times R2. You can see both top and bottom with multiplying by I. So we can cancel the I's. That gives me the result that I want. V1 divided by V2 is equal to R1 divided by R2. So in a potential divider circuit, the ratio of the voltages is equal to the ratio of the resistances. Let's use that to solve this problem. I've got a resistor with 10 ohms, with 6 volts across it, and an unknown resistor with 9 volts across it, in a series circuit as a potential divider. Here's the formula that I'm going to use. V1 over V2 is equal to R1 over R2. V1 is 9 volts, and V2 is 6 volts. R1 is the unknown, and R2 is 10 ohms. Let's rearrange that. R1 equals 9 divided by 6 multiplied by 10, which equals 15 ohms.