# Question Video: Finding the Resistance of a Resistor in a Potential Divider

A potential divider has an input voltage of 48 V. The resistance of the second resistor, πβ, is 100 kΞ©. The output voltage is drawn across the second resistor, πβ. What resistance must the first resistor, πβ, have in order to produce an output voltage of 32 V?

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### Video Transcript

A potential divider has an input voltage of 48 volts. The resistance of the second resistor, π two, is 100 kiloohms. The output voltage is drawn across the second resistor, π two. What resistance must the first resistor π one have in order to produce an output voltage of 32 volts?

Okay, so in this case, weβre dealing with a potential divider circuit. So letβs first start by drawing our potential divider circuit. Letβs first start by recalling that a potential divider circuit consists of two resistors connected in series. Weβll call the resistance of our first resistor π one and the resistance of our second resistor π two, as weβve been told to in the question. We also recall that across this part of the circuit, some sort of power source is connected, often a battery, but not necessarily so. And this power source is what provides whatβs known as the input voltage. Letβs call this input voltage π subscript in. And in this case, in the question, weβve been told that this input voltage is equal to 48 volts.

Now, additionally, weβve been told that the resistance of the second resistor π two is 100 kiloohms. Thatβs 100000 ohms because the prefix kilo- means 1000. Weβve also been told that the output voltage in our potential divider circuit is drawn across the second resistor. What this essentially means is that we connect a pair of wires here and here so that we can connect some components in parallel with our resistor π two. In this case, weβre not really worried about which components are connected here, but just that the output voltage, which we will call π subscript out, is drawn across our second resistor.

Now, the reason that we know that this is the output voltage is because the voltage across our second resistor must be the same as π out. The reason for this is that our resistor π two and whatever components are connected here are connected in parallel. And components in parallel have the same potential difference across them. Therefore, to reiterate, the potential difference across resistor π two is π subscript out. Thatβs the output voltage. And thatβs also the potential difference across whatever components are connected here and is therefore the output voltage.

Weβve been told in the question that the output voltage must be equal to 32 volts. And weβve been asked to try and work out what the resistance of the first resistor π one must be in order for this to be true. So in order to answer this question, we will need to recall the potential divider equation. This equation tells us that the potential difference across resistor π two, which weβre calling π two, is equal to the resistance π two divided by the sum of the resistances π one plus π two all multiplied by the input potential difference π subscript in.

Now remember, the potential difference across our second resistor π two is the same as our output potential difference because theyβre connected in parallel. So we can replace π two with π out here, at which point we see weβve already got a value for π out, a value for π two, and a value for π in. We just need to rearrange to solve for π one. We can do this by multiplying both sides of the equation by π one plus π two and dividing both sides of the equation by π out. Because this way on the left-hand side weβve got π out over π out, which is equal to one. And on the right-hand side, weβve got π one plus π two both in the numerator and denominator.

Then we simply subtract π two from both sides of the equation so that weβre left with π one on the left, at which point we simply substitute in the values on the right-hand side, taking care to notice that in this fraction the unit of volts is both in the numerator and denominator and that if weβre going to stick with kiloohms, then our final answer for π one is going to be in kiloohms as well. When we simplify all of the right-hand side, we find that our final answer is π one is equal to 50 kiloohms.