Solving the Missing Resistor Problem in a Parallel Circuit

[Hazel Appraiasal]

Summary:: How de you find the missing resistor (R6) in this problem. I assume you take the current flow of the parallel circuit to find the resistance of R6 but I feel like there is still some missing information

 

[berkeman]

Looks like a straightforward series-parallel resistor combination problem to me. Can you post your work?

 

[FactChecker]

Although the value of $R_6$ is not given, the voltage across $R_9$ is given. That additional information of the voltage drop should give enough facts to solve the problem.

 

[Hazel Appraiasal]

This is all I've done. I figure that if I add all of the resistances of the series circuit then find the current, I could then use that current to solve the parallel circuit. I got something around 0.01 amp so I figure that the total current of the parallel circuit must be 0.01 amp and then I could use that to find the resistance of R6

 

[FactChecker]

You can get the current in the main loop directly from the voltage drop at $R_6$.
0.01 A is correct. I don't know any other way that you might have used.
Given that, can you write down an equation leaving $R_6$ undefined and solve for $R_6$?
(You can combine all the series resisters in the main loop, combine the parallel $R_5$ and $R_7$ but leave $R_6$ as a variable. Then solve for $R_6$)

 

[Hazel Appraiasal]

You can get the current in the main loop directly from the voltage drop at $R_6$.
0.01 A is correct. I don't know any other way that you might have used.
Given that, can you write down an equation leaving $R_6$ undefined and solve for $R_6$?
(You can combine all the series resisters in the main loop, combine the parallel $R_5$ and $R_7$ but leave $R_6$ as a variable. Then solve for $R_6$)

I don't think so because the total current of the parallel circuit must be 0.01 amps which if I calculated correctly results in a total resistance of the parallel circuit to be 10 000 and then I can subtract resistances from there.

 

[FactChecker]

I don't think so because the total current of the parallel circuit must be 0.01 amps

Yes, you can get that directly from the 5-volt drop across the $500 \Omega$ $R_9$ resister.

which if I calculated correctly results in a total resistance of the parallel circuit to be 10 000 and then I can subtract resistances from there.

You can not get a larger combined parallel resistance than the least of the individual parallel resisters. So that must be no more than $3.9 k\Omega$
I guess that you mean the resistance of the entire circuit. I agree.