Finding Internal Resistance of a Circuit

  • #1
Josh225
51
3

Homework Statement


See image attached.

I am currently stuck on part b.

Homework Equations


(Rt) (Rint) / Rt + (Rint)

The Attempt at a Solution


I thought that Rt would be 26.7 kohms, but my answer is not matching up with the answer in the book. I am unsure why since the resistors are in series, so I figured I would just add the 2 values. Once I find that, I just plug in total resistance and internal resistance into the formula above. Then I would have to convert Mohms to kohms.
Does that sound right?

Thank you in advance
 

Attachments

  • 20160728_150206.jpg
    20160728_150206.jpg
    36.5 KB · Views: 427
Physics news on Phys.org
  • #2
Can I suggest you make a sketch circuit showing the DMM modeled as an ideal DMM and an 11M resistor.

You don't say what Rt is?
You mention resistors in series but don't say which ones you are talking about.
 
  • #3
CWatters said:
Can I suggest you make a sketch circuit showing the DMM modeled as an ideal DMM and an 11M resistor.

You don't say what Rt is?
You mention resistors in series but don't say which ones you are talking about.
Im not 100% sure by what you mean by making a sketch. I thought 26.7 kilohms would be the total resistance within the circuit (by adding the 2 resistors together). If not, how would you do so?

Here is my sketch, but am unsure it this is what you were suggesting.
 

Attachments

  • 20160728_191621-1.jpg
    20160728_191621-1.jpg
    26.9 KB · Views: 396
  • #4
Josh225 said:
Im not 100% sure by what you mean by making a sketch. I thought 26.7 kilohms would be the total resistance within the circuit (by adding the 2 resistors together). If not, how would you do so?

Here is my sketch, but am unsure it this is what you were suggesting.
Part (b) of the question asks you to measure V2 assuming an internal impedance of 11M in the meter. Your sketch shows the meter across the wrong resistor.

I would suggest you calculate the current in the circuit in the usual way, then calculate the voltage (theoretically) across R2. Once you have done that, measuring the voltage with the meter effectively places two resistors in parallel. So what do you think would happen to the circuit current if you change the effective resistance of R2? And if the current is different, what would happen to the voltage drop across R2, would it be the same as you calculated in part (a)?
 
  • #5
Yes your sketch is exactly what I meant but as Numbskull said you have the meter across the wrong resistor.
 

Similar threads

Back
Top