Question about deriving Resistance from data on Electric Current in Two Different Scenarios

I am preparing for an undergraduate degree in Physics, with no teacher in sight until October. The textbook I am using to prepare myself, gave me the following exercise about electric current and left me beyond confused. I would really appreciate it if somebody could help out.

SCENARIO ONE: The Ammeter reads: 0.48103 Amp
The Voltmeter reads: 119.7959 Volt

SCENARIO TWO: The Ammeter reads: 0.47904 Amp
The Voltmeter reads: 120 Volt

The voltage between a and c is identical in both scenarios.

The book asks me to calculate the resistance of the device (R), the resistance of the Ammeter (A) and the resistance of the Voltmeter (V).

The following is my reasoning, and I am dying to find out where my mistake is:
-The difference in voltage between the two readings shows that the tension between b and c is 0.2405 Volt.
-The current shown by the Ammeter in the first scenario is the total current for the system, whereas the current shown by the Ammeter in the second scenario is the current for its own branch only. Therefore the current of the Voltmeter's branch is the total current minus the current of the Ammeter's branch: 0.48103-0.47904 = 0.00199 Ampère.
-The resistance of the device itself can be calculated: the current of its 'branch' is 0.47904, and as I=V/R and V = 119.7595 between a and b, the resistance of the device itself is 250 Ω.

Here I get stuck, and I need to know where my logic fails:
-Regarding the resistance of the Voltmeter, I have the current (0.00199 Ampere), but I have two different voltages, and they seem equally valid to be plugged into the equation I=V/R. In that sense, I would get two different results from the equation R=V/I. Now, resistance is dependent on ρ, L and A, and one could argue that the L changed here. Either way, I am left with the values R = 60301 Ω and R = 601380 Ω.

-Regarding the resistance of the Ammeter, I have the voltage for b-c (0.2405), but I need the current. Here, there are two options for the Current, but - again - they lead to different values for the Resistance (R = 0.5 Ω and 0.502 Ω respectively). However, in this case, there is no logical reason why the resistance would have changed, because there is no reason for ρ, L and A to have changed. Factually speaking, the Voltage is identical in both cases (0.2405), yet the current has changed: looking at I=V/R, that must mean that the resistance has changed, but it makes no sense…

Thank you!

• You have 3 unknowns (variables) but only 2 knowns (constraints/equations) - this is not enough. It looks as though this cannot be solved. Which may explain why you are confused. – sammy gerbil Aug 27 '18 at 21:04
• @ Pregunto : your calculation is wrong "the so called tension(i., potential difference) between b and c is $120-119.7959=0.2041 (not \mathbf {0.2405})$" – Faraday Pathak Aug 28 '18 at 10:20
• @sammy gerbil : it is solvable because node $\mathbf a$ is always $\mathbf { tied}$ to same potential of $120$ with repsect to $\mathbf {ground }\ \ \$ node(i.e, c) , ........irrespective of which scenario you look – Faraday Pathak Aug 28 '18 at 10:23

take c as ground node

i assign potentials to node $\mathbf a$ and node $\mathbf b$ as $\mathbf V_{a}$ and $\mathbf V_{b}$ respectively with repect to ground node i.e, c

then, $\mathbf{ V_{a}=120}$ volt (in both from scenarios)

but , $\mathbf {V_{a}-V_{b}}=119.7959$ volts (from scenario 1)

$\implies \mathbf V_{b}=120-119.7959=0.2041$ volts= voltage across ammeter

but ammeter draws current of $0.47904 A$

= resistance of ammeter is $\mathbf {R_{A}} =\dfrac{0.2041}{0.47904}\approx 0.4260\ \Omega \tag{1}$

also,

resistance $\mathbf R=\dfrac{120-0.2041}{0.47904}\approx 250.075\ \Omega\tag{2}$

resistance of voltmeter $\mathbf R_{V}=\dfrac{119.7959}{0.48103-0.47904}\approx 60198.945\ \Omega \tag{3}$