Value of the shunt resistor in this circuit

[Bolter]

Homework Statement

Find value of shunt resistor in circuit

Relevant Equations

Ohms law

Here are a couple of questions that I have been trying to answer and had a go at it.
Not sure if I answered some incorrectly so was hoping to get some guidance

For this one I have tried to make use of Kirchhoff's 2nd law to help me, but this is what I have ended up with when using it

Thanks!

 

[haruspex]

Your work on the first three looks fine.
In the last one, isn't Ic the current through the ammeter? That is not 1A.
Seems to me that the set up is the same as in the third problem, so you can use the same method.

 

[gneill]

Moderator's note: Please post only a single problem per thread. This post should have been broken up into four separate threads.

 

[Bolter]

Your work on the first three looks fine.
In the last one, isn't Ic the current through the ammeter? That is not 1A.
Seems to me that the set up is the same as in the third problem, so you can use the same method.

Thanks I have realized that voltage across the shunt and coil would be the same as they are connected in parallel. And that current would also split too at the junction.

I did this and get a resistance of 0.25 ohms for the shunt

 

[Bolter]

Moderator's note: Please post only a single problem per thread. This post should have been broken up into four separate threads.

My apologies, I'll try not to do this again next time

 

[haruspex]

Thanks I have realized that voltage across the shunt and coil would be the same as they are connected in parallel. And that current would also split too at the junction.

I did this and get a resistance of 0.25 ohms for the shunt

View attachment 256866

Yes, but you can get there more quickly. You know that 1A-200mA=800mA is going through the shunt. That's 4 times the current for the same voltage, so a quarter of the resistance.

 

[Bolter]

Yes, but you can get there more quickly. You know that 1A-200mA=800mA is going through the shunt. That's 4 times the current for the same voltage, so a quarter of the resistance.

Yes this is much more efficient way of doing now that I can see :)