Kirchhoff's Rules Homework: Current & Voltage Measurements

[lha08]

Homework Statement

So i had to do an experiment which consisted of measuring the current and voltage across different resistors. Here's my question: i recorded the current through the wire that passes through the resistor and compared it with the current which i calculated (theoretical). In this case, when i apply the junction rule using the current i read from the ammeter, it is NOT equal to zero (somehow its higher than 0). Like why is this the case? I'm not sure if its right but could it be that we were not using an ideal circuit, that is, there is resistance in the wires that could have affected our ammeter readings? Like I'm thinking that the resistance in the wire also affect our voltmeter readings as well...am i on the right track?

Homework Equations

The Attempt at a Solution

 

[xcvxcvvc]

Homework Statement

So i had to do an experiment which consisted of measuring the current and voltage across different resistors. Here's my question: i recorded the current through the wire that passes through the resistor and compared it with the current which i calculated (theoretical). In this case, when i apply the junction rule using the current i read from the ammeter, it is NOT equal to zero (somehow its higher than 0). Like why is this the case? I'm not sure if its right but could it be that we were not using an ideal circuit, that is, there is resistance in the wires that could have affected our ammeter readings? Like I'm thinking that the resistance in the wire also affect our voltmeter readings as well...am i on the right track?

Homework Equations

The Attempt at a Solution

Are you saying that the sum of the current entering and leaving a node measured during a lab equaled a nonzero?

 

[lha08]

Are you saying that the sum of the current entering and leaving a node measured during a lab equaled a nonzero?

when i calculated it i got like 0.07 A, which is obviously not equal to zero and does not obey the Junction Rule..and i can't figure out what particular reasons why this is..

 

[xcvxcvvc]

when i calculated it i got like 0.07 A, which is obviously not equal to zero and does not obey the Junction Rule..and i can't figure out what particular reasons why this is..

Hmm. Do you remember if during measurement the ammeter's readings fluctuated rapidly (such as +/- 50 mA)? Maybe your equipment's precision prevented the demonstration of junction rule.

 

[lha08]

Hmm. Do you remember if during measurement the ammeter's readings fluctuated rapidly (such as +/- 50 mA)? Maybe your equipment's precision prevented the demonstration of junction rule.

They were fluctuating but not that quickly but that is a possibility. But like generally do ammeter and voltmeter equipment contain resistance or something that would cause discrepancies in their readings? But like i said earlier, could the connecting wires that i used also cause discrepancies because they have resistance? ..not really sure

 

[xcvxcvvc]

They were fluctuating but not that quickly but that is a possibility. But like generally do ammeter and voltmeter equipment contain resistance or something that would cause discrepancies in their readings? But like i said earlier, could the connecting wires that i used also cause discrepancies because they have resistance? ..not really sure

Ammeters have fractions of an ohm, so unless you were using relatively tiny resistances in your circuit (meaning those fractions of an ohm were significant), the ammeter's resistance doubtfully worsened your results.

With these types of measurements, usually there isn't much room for the theory to go wrong -- you have a flow of charge inward that must equal the flow of charge outward, else charge was created or destroyed. Yes, the actual current values will be different compared to the theoretical calculations (mostly due to the resistors' tolerances, wire resistance, etc.), but still, the sums leaving and entering should equal zero or a very small value due to rounding error/precision of the measuring device. 70mA seems high, though.