Question

When an unknown resistance $R_x$ is placed in a Wheatstone bridge, it is possible to balance the bridge by adjusting $R_3$ to be $2500 \Omega$. What is $R_x$ if $\dfrac{R_2}{R_1} = 0.625$?

Final Answer

$1.56 \textrm{ k}\Omega$

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Video Transcript

This is College Physics Answers with Shaun Dychko.
In a Wheatstone bridge, this resistance number three is adjusted such that no current will want to flow through this galvanometer. I don't know why they put this switch in there because this switch is closed actually, and despite being closed no current flow through because this resistance is adjusted so that the voltage across it is equal to the voltage across resistance one. So that's what this is saying algebraically. This is saying voltage from point A to B is the same as voltage from A to D. So the current through resistor three which is

*I two*, multiplied by*R three*, is equal to the current to resistor*R one*which is*I one*multiplied together. Likewise, since there is no current flowing through this galvanometer, the voltage across this unknown resistance*R x*is going to equal the voltage across resistor*R two*. So we say*I two R x*equals*I one R two*. Then we can take the ratio of this to that and this to that. We can divide these two equations in other words, and we'll be able to cancel*I two*and*I one*which we don't know and now we don't care about them because they cancel away. We get R unknown,*R x*, divided by*R three*equals*R two*over*R one*. Then we can solve for*R x*by multiplying both sides by*R three*. So*R x*is*R three*times*R two*over*R one*and we don't know what*R two*and*R one*is but we're told what the ratio of the two are and that's what we care about. We're told that*R three*is 2500 ohms. So 2500 times 0.625 is 1.56 kilo-ohms, that is the unknown resistance*R x*.