Navigate the Res-Monster Maze: Find R's Current

[L_landau]

Homework Statement

Res-monster maze. In Fig. 27-21, all the resistors have a resistance of 4.0 and all the (ideal) batteries have an emf of 4.0 V. What is the current through resistor R? (If you can find the proper loop through this maze, you can answer the question with a few seconds of mental calculation.)

Homework Equations

V = IR

The Attempt at a Solution

Can't I just use the loop rule which says that the sum of all the voltage drops in a closed loop is equal to zero? So I would just go around the smallest loop that encompasses the R, and I'd have
ε - IR - IR = 0 -> I = ε/2R but this isn't correct.

 

[magoo]

No, you can't do that. For the loop that you selected, the current (I) is shared by other loops.

 

[L_landau]

But if I don't go through any other resistors besides R, and take a circle around the circuit back to R (going through a few potentials), that path doesn't share it's current with other loops? What's the thought process for that?

 

[magoo]

Can you draw the loop you selected on the drawing?

In the problem statement, all the resistors have a value of 4 ohms. I would label them all r. The unknown R could have some other value.

Is your equation involved with both r plus R? Please show a sketch.

 

[phinds]

But if I don't go through any other resistors besides R, and take a circle around the circuit back to R (going through a few potentials), that path doesn't share it's current with other loops? What's the thought process for that?

Uh ... you really need to rethink that.

 

[ehild]

But if I don't go through any other resistors besides R, and take a circle around the circuit back to R (going through a few potentials), that path doesn't share it's current with other loops? What's the thought process for that?

You can take the potential zero at point O. What is the potential at A? Find a path from O to A going through batteries only.

 

[ehild]

But if I don't go through any other resistors besides R, and take a circle around the circuit back to R (going through a few potentials), that path doesn't share it's current with other loops? What's the thought process for that?

The potential difference across an ideal battery stays the same if you connect anything to its terminals. If you know the potential difference across R you can get the current through it.