Resistors in Series, in Parallel

[rabcdred]

Homework Statement

See the attached image.

Homework Equations

V=IR, Kirchoff's Law

The Attempt at a Solution

The voltage drop across resistors in parallel are equivalent due to Kirchoffs law (at least I think so?), so V=R(eq,top)I(top)=R(eq,bottom)I(bottom)--> I(top)(52)=I(bottom)(7+Rx)

For each resistor, an equation using Ohm's Law:
V1=I(top)(14)
V2=I(top)(38)
V3=I(bottom)(7)
V4=I(bottom)(Rx)

As the ammeter reads zero, I thought the voltage drop across resistor 1 and 3 were equal, so V1=V3. Rearranging the equation and substituting in the top equation-->
(V/14)(52)=(V/7)(7+Rx), which yields Rx=19.

I don't think this is right though. Please help! Thanks.

 

[PeterO]

Homework Statement

See the attached image.

Homework Equations

V=IR, Kirchoff's Law

The Attempt at a Solution

The voltage drop across resistors in parallel are equivalent due to Kirchoffs law (at least I think so?), so V=R(eq,top)I(top)=R(eq,bottom)I(bottom)--> I(top)(52)=I(bottom)(7+Rx)

For each resistor, an equation using Ohm's Law:
V1=I(top)(14)
V2=I(top)(38)
V3=I(bottom)(7)
V4=I(bottom)(Rx)

As the ammeter reads zero, I thought the voltage drop across resistor 1 and 3 were equal, so V1=V3. Rearranging the equation and substituting in the top equation-->
(V/14)(52)=(V/7)(7+Rx), which yields Rx=19.

I don't think this is right though. Please help! Thanks.

19Ω is a perfect answer - so probably your reasoning is correct too as answers like 19 don't usually appear by co-incidence.