- #1
mikebc
- 20
- 0
Hi,
I am having a problem with my lab report and am hoping somebody can help me understand it better.
Here is my problem:
I set up a series circuit with 5 resistors of 427 ohms each. The voltage of the battery was 8.70V to start. Once the battery was removed from the circuit its reading increased to 8.92V. In the experiment I am supposed to consider the battery to be an extra resistor and determine what its resistance would be so that all the voltages measured before would add up to the battery voltage if the drop across this battery was added in.
This is how I calculated it out but I do not think that this is the right way to do it and cannot think of another way to do it. I worked out the current to be 0.021amps (I=V/R 8.92/427=0.021) and then I worked out the resistance using the difference between the starting battery voltage and the final voltage (R=V/I 0.8/0.021=38ohms).
So my answer would be that the battery would have to have a resistance of 38Ohms.
As an additional note, I don't know if this is needed to be known but the voltage through each resistor was 8.7, 6.95, 5.22, 3.47, and 1.44 respectively.
Any help would be extremely appreciated, thank you!
I am having a problem with my lab report and am hoping somebody can help me understand it better.
Here is my problem:
I set up a series circuit with 5 resistors of 427 ohms each. The voltage of the battery was 8.70V to start. Once the battery was removed from the circuit its reading increased to 8.92V. In the experiment I am supposed to consider the battery to be an extra resistor and determine what its resistance would be so that all the voltages measured before would add up to the battery voltage if the drop across this battery was added in.
This is how I calculated it out but I do not think that this is the right way to do it and cannot think of another way to do it. I worked out the current to be 0.021amps (I=V/R 8.92/427=0.021) and then I worked out the resistance using the difference between the starting battery voltage and the final voltage (R=V/I 0.8/0.021=38ohms).
So my answer would be that the battery would have to have a resistance of 38Ohms.
As an additional note, I don't know if this is needed to be known but the voltage through each resistor was 8.7, 6.95, 5.22, 3.47, and 1.44 respectively.
Any help would be extremely appreciated, thank you!