In Lab 02 Investigation 2 you have the battery and the current in the circuit wh

Q: In Lab 02 Investigation 2 you have the battery and the current in the circuit wh

with only one bulb I1 = V/R1 = V/R V1 = I1*R = V with two bulbs I2 = V/(R1+R2) = V/2R V1 = I2*R1 = V/2R*R = V/2 3. No, the statement in this problem is incorrect. Since in case of two bulbs in series the overall current is lower than in case of a single bulb, therefore the voltage drop across the internal resistance of the battery is also lower and we observed that the voltage across the battery was slightly higher.

ID: 1512189 • Letter: I

Question

In Lab 02 Investigation 2 you have the battery and the current in the circuit when 1 and then 2 bulbs in series were connected to the same battery. If you did everything correctly you probably found that the average current through the battery with 2 bulbs was lower (as expected), but the average voltage across the 2 bulbs was actually slightly higher (unexpected) than in the case with only one bulb connected.Approximating the real battery as an ideal voltage source with some internal resistance can you explain your observation? No, there is not enough data collected in this experiment to explain the phenomenon. Yes, since in case of two bulbs in series the overall current is higher than in case of a single bulb. Therefore the voltage drop across the internal resistance of the battery and thus across the bulbs is also higher. No, the statement in this problem is incorrect. Since in case of two bulbs in series the overall current is lower than in case of a single bulb, therefore the voltage drop across the internal resistance of the battery is also lower and we observed that the voltage across the battery was slightly higher. Yes, since incase of two bulbs in series the overall current is lower than in case of a single bulb. Therefore the voltage drop across the internal resistance of the battery is also lower and thus the voltage drop across the bulbs is higher.

Explanation / Answer

with only one bulb

I1 = V/R1 = V/R

V1 = I1*R = V

with two bulbs

I2 = V/(R1+R2) = V/2R

V1 = I2*R1 = V/2R*R = V/2

3. No, the statement in this problem is incorrect.

Since in case of two bulbs in series the overall current is lower than in case of a single bulb,

therefore the voltage drop across the internal resistance of the battery is also lower and

we observed that the voltage across the battery was slightly higher.

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In Lab 02 you have measured the voltage V across the battery and the current I i

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In Lab 02 Investigation 2 you have the battery and the current in the circuit wh

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