Finding equivalent resistance in a multi-branch circuit

[naftacher]

Homework Statement

When the ammeter reads 7 A, the voltmeter reads

Relevant Equations

I would assume this requires simple V=iR (ohms) and likely Kirchoff's Laws surrounding currents.

I am thinking that this circuit needs to be simplified somehow? Finding an equivalent resistance and moving from there?

 

[robphy]

One can try simplification.
But if that doesn't work, one could try applying Kirchhoff's Laws

 

[berkeman]

Homework Statement:: When the ammeter reads 7 A, the voltmeter reads
Relevant Equations:: I would assume this requires simple V=iR (ohms) and likely Kirchoff's Laws surrounding currents.

View attachment 265636
I am thinking that this circuit needs to be simplified somehow? Finding an equivalent resistance and moving from there?

What is the impedance of an ideal voltmeter? So how much current gets diverted through it? Use that fact to simplify the resistor circuit to find the overall voltage, then break it back into the two branches to calculate the voltage dividers, etc.

 

[etotheipi]

At a junction like the one in your diagram, current will split in proportion to the conductivities, so $i_A = \frac{R_B}{R_A + R_B}i$

Spoiler: Try to prove it yourself first!

$$i_A R_A = i_B R_B \implies i_B = \frac{R_A}{R_B} i_A$$and as such we have$$i = i_A + i_B = i_A \left (1 + \frac{R_A}{R_B} \right) = i_A \left(\frac{R_A + R_B}{R_B} \right)$$so the current flowing down the branch with resistance $R_A$ is$$i_A = \frac{R_B}{R_A + R_B}i$$

 

[naftacher]

At a junction like the one in your diagram, current will split in proportion to the conductivities, so $i_1 = \frac{R_2}{R_1 + R_2}i$

Spoiler: Try to prove it yourself first!

$$i_1 R_1 = i_2 R_2 \implies i_2 = \frac{R_1}{R_2} i_1$$and as such we have$$i = i_1 + i_2 = i_1 \left (1 + \frac{R_1}{R_2} \right) = i_1 \left(\frac{R_1 + R_2}{R_2} \right)$$so the current flowing down the branch with resistance $R_1$ is$$i_1 = \frac{R_2}{R_1 + R_2}i$$

I have seen this equation before. How am I to extend this to R3 and R4? or do I not need to

 

[etotheipi]

I have seen this equation before. How am I to extend this to R3 and R4? or do I not need to

Whoops, don't take the numbers too literally, I've changed them to A's and B's to make it more general. It's applying to cases where you have multiple branches of different resistance between two nodes. $R_A$ and $R_B$ are the total resistances of those branches.

@berkeman gave you an important hint about the voltmeter, to explain why we can use the reasoning we did to get that result...