series and parallel circuits

The last several labs for the quarter involve series and parallel circuits.
While we did an asisgnment on Crocodile Physics where we examined resistors and found the patterns and equations for equivalent resistances, we expanded that in the lab the last few days by looking at how the current and voltage were distributed around a circuit.

The basic logic is this:
1) in a series circuit, the current is the same throughout the series.
2) in a parallel circuit, each bulb is connected independently to the battery, so the voltage drop is the same across each bulb.

More details:

1) in a series circuit, the current that comes out of the battery must go into the first bulb, then out and then into the next and then out, and then into the next and then out... etc. Since the current doesn't "fade" or "get used up", that means that the current is the same value through each resistor in the series circuit.
The voltage, however, drops across each bulb so that the total voltage change should equal the voltage imposed on the circuit by the power supply.
The amount of voltage change for each bulb depends upon how much resistance the bulb has (V=IR). The more resistance, the more "oomph" that goes into going through. This helps us figure out the power, too. P=IV. The total energy that is used up in "pushing" charges through the resistor is the amount of heat energy that causes the bulb to light up. Thus, the brighter the bulb, the more power.

2) in a parallel circuit, the current comes out of the battery and splits up. some goes one way, some goes the other. Since each bulb is connected directly to the battery, the voltage change across each bulb is the total voltage change that the battery imposes. No sharing.

In other words...
in a series circuit, voltages add up and current is constant
in a parallel circuit, voltages are the same, and current adds up.

So... what about that dang combination circuit?
Well, you can first of all consider it as two "things" in series. The first is a single resistor. The second "thing" is a pair of resistors in parallel. So, if they are all identical bulbs, then the second "thing" has a resistance that is half that of the first bulb. That means that more energy is being used up pushing lots of charges through that first bulb. That's the one that will glow brightly. Then by the time you get to the two in parallel, the charges can split up and go both ways, so it is pretty easy going. Half the current goes through each, so they barely will light up, if at all.

Let's do a sample problem of the combination circuit. It won't help a lot to just read this. You have to try to figure it out and be able to do these calculations on your own...

This circuit is a single resistor connected to a pair of resistors in parallel. Like we've seen in class several times already.

All resistors with a resistance of 6 ohms, hooked up to a 4V battery.

The total resistance of the circuit is 6 + 3 = 9 ohms, since the two in parallel combine to make 3.

Using V=IR, we have 4=I*9, so I=4/9 = 0.44 amps. That is the current leaving the battery and entering the whole circuit.

So, that's the current entering the first resistor.
The first resistor remember has a resistance of R=6 ohms, and with a current of I=0.444 Amps, then we know that the Voltage change is V=IR = 0.444*6 = 2.66 V

So, then we know that the voltage change across the parallel chunk of the circuit must be whatever is leftover.
V= 4 V - 2.66 V = 1.33 Volts. So 1.33 Volts is the difference across each of the two resistors in parallel.
Since each resistor in the parallel chunk has a resistance of 6 ohms, and a voltage change of 1.33 Volts, then we know, from V=IR, that the current through each resistor must be
I=V/R=1.33/6 = .22 Amps.
But of course we knew this already, since the total current through the circuit is .44 amps, and the since the resistors are the same, the current should split evenly.

The power output of each bulb is:
#1 P=IV = 0.44 amps * 2.66 Volts = 1.17 Watts
#2 P=IV = 0.22 amps * 1.33 Volts = .293 Watts
#3 P=IV = 0.22 amps * 1.33 Volts = .293 Watts

Notice that the single bulb gets 4 times the energy every second. So it is very bright, the others are very not bright. Half the current AND half the voltage makes it pretty dim.