III. Using voltage in series circuits Use CCK to build the clrcults below with a
ID: 2304080 • Letter: I
Question
III. Using voltage in series circuits Use CCK to build the clrcults below with a battery at about 12 volts and light blbs. Tum on the voltmeter and ammeter to measure voltage of the battery and current into it. Record bulb brightness with descriptive language. Figure 1 Figure 2 Figure 3 attery Current voltage into of bulbs A) a. Summarize the relationships you observed and explain what you think is happening Test to see if changing the battery voltage causes you to modify any of your conclusions. measured and any conclusions you draw from your tests. What happens when you take a wire out of a circuit? Explain what you think is happening Test using the voltmeter or ammeter in different ways. For example: Does it matter if you take the reading on the lett or right of the battery? Switch the meter ends? Describe your tests and results. b. Explain what you c. d. 11/3/2008 Loeblein Some Properties of Electric Circuits (Uses CCK only) IV. Using voltage in parallel circuits Redo Part III but use igures 4-6 for the circuits Make a new table and answer the questions. Figure 4 Figure 5 Figure 6Explanation / Answer
Assuming we use identical bulbs in the experiment.
The light bulb glows according to the power output. P = VI where V is the voltage across its terminals and I the current that flows through it.
III. In the first configuration, for a voltage of 12V, say we get I1 current across the bulb. (Hence resistance of bulb = 12/I1). And power will be 12I1.
In the second configuration, there are two bulbs in series hence their resistances add up to give net 24/I1 and current across both the bulbs are 12/(24/I1) = I1/2. And voltage across each bulb will be equally divided as 6V each. Hence power across each bulb will be 6I1/2 = 3I2, which is less than the first configuration, indicating lesser brightness.
Similarly in the third configuration, current will be 12/(36/I1) = I1/3, the voltage across each bulb is 4V and power across each bulb 1.6I1, indicating even dimmer bulb.
IV. In the first configuration, as in III, for a voltage of 12V, assume we get I1 current across the bulb. (Hence resistance of bulb = 12/I1). And power will be 12I1.
In the second configuration, there are two bulbs in parallel hence their net resistances will be (12/I112/I1)/(12/I1+12/I1) = 6/I1 and voltage across both the bulbs are 12V. Net current is 12/(6/I1) = 2I1. And current across each bulb will be equally divided as I1 each. Hence power across each bulb will be 12I1, which is equal to the first configuration, indicating same brightness.
Similarly in the third configuration, the voltage across each bulb = 12V, net current will be 12/(4/I1) = 3I1, the current across each bulb is I1 and power across each bulb 12I1, indicating same brightness.