# Series circuit lab problem: voltage drop?

1. Nov 6, 2007

### mikebc

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!

2. Nov 6, 2007

### rl.bhat

There are 5 resistances each of 427ohm So the total resistance in the external circuit = 5x427=2135 ohm. Voltage across external resistance = 8.70 = 8.92x2135/(2135 + r) where r is the battery resistance. Find the value of r.

3. Nov 6, 2007

### mikebc

Thanks for the quick reply. I see one mistake that I did by not adding up the resistances but I still don't quite understand this equation.

4. Nov 6, 2007

### Staff: Mentor

When you open-circuit the battery and measure the voltage, that is the source voltage with no current flowing, so there is no voltage drop across the internal resistance of the battery, right?

And then when you connect the resistor string outside, a current flows through the internal battery resistance, which causes a voltage drop, which gives you a lower battery voltage measured at the battery terminals. The internal drop depends on the current, right? More current thought the external circuit means you get a bigger voltage drop across the battery's internal resistance, so you measure a smaller external battery voltage at the battery terminals.

Now, you know how much current is flowing in the external circuit with the resistors connected, because you know the total voltage drop across the external resistors, and you know their total resistance. What is the total external current?

And knowing that current, and how much the battery voltage decreases when the resistors are connected, tells you what.....?

5. Nov 6, 2007

### Ouabache

You may have noticed there is more than one way to find Ri (battery's internal resistance). You can use voltage division as rl.bhat suggests or using ohm's law twice as berkeman has outlined. Both are equivalent and may be used to double-check your solution.

6. Nov 6, 2007

### mikebc

You guys are awesome for explaining this to me. I just want to let you know that I am still processing the information, it hasn't quite clicked with me yet. From what I understand the external circuit is I=V/R = 8.70/2135 = 0.0041 amps. So the internal resistance is R=V/I = 8.92/0.0041 = 2176 ohms. Then the difference between 2176 and 2135 would be the resistance for the battery? Answer 41 ohms. Does this make sense or am I still not correctly understanding? Thanks for your patience.

7. Nov 7, 2007

### Staff: Mentor

I didn't check your math, but it sounds like you are on the right track.

8. Nov 7, 2007

### mikebc

Thanks for letting me know that I am on the right track, berkeman, and thank you to everybody else that helped me to get on the right track:)

9. Nov 7, 2007

### rl.bhat

Internal resistance