Using a multimeter as an ammeter

[TomHart]

I wasn't too confident about it at first but then we started answering their questions. I'll just lay them out for you.

i) Connect the positive terminal of the battery and set the current to 16 mA. What is the voltage V1 across the component M?

We went along with 0.16 A instead of 16 mA and we got a voltage of 3.16 V.

ii) Reverse the component M and set the current to 32 mA. What is the voltage V2 across the component M?

We went along with 0.32 A and we got a voltage of 2.52 V.

iii) Component M contains the items shown in the diagram shown below (the schematic attached previously with the diode in series with one resistor and in parallel to another resistor). Use part (i) to find the resistance of the resistor R'.

This was the reverse biased connection so by theory if we assume that the diode did not allow any current to pass through then all the current went to the resistor in parallel and so R' = V/I = 3.16 V / 0.16 A = 19.75 ohms which is close since that resistor was 20 ohms.

iv) Assume the voltage drop across the diode is 0.60 V when forward biased. For the arrangement in part (ii) when the upper and lower part of component are conducting, find the following:

-current flowing through resistor R'

I = V/R = 2.52 / 19.75 = 0.127 A

- current flowing through resistor R

I = 0.32 A - 0.127 A = 0.193 A

-the resistance of the resistor R.

The voltage across R is 2.52 V -0.60 V=1.92 V and therefore R = 1.92 V / 0.193 A = 9.9 ohms which is very close to 10 ohms (the original value we were using for R)

Looks great - exactly what you would expect!

 

[Taniaz]

Looks great - exactly what you would expect!

Thank you for your help!