Electrical Circuit Question

Electrical Circuits Questions?

1. Calculate the current in a 44 V battery that powers a pair of 25 Ω resistors connected in series.

2. Calculate the current in a 38 V battery that powers a pair of 15 Ω resistors connected in parallel.

3. The rear window defrosters on automobiles are made up of several strips of heater wire connected in parallel. Consider the case of four wires, each of 6 Ω resistances, connected to 24 V.

(a) What is the equivalent resistance of the four wires? (Consider the wires to be two groups of two.)

(b) What is the total current drawn?

A question on electrical circuits?

The circuit diagram below shows a cell connected in series to an ammeter and a box which contains two other electrical components. A voltmeter is connected in parallel with the box.

http://img23.imageshack.us/img23/2901/circuitk.jpg

When the connections to the box are reversed, the ammeter reading decreases but the voltmeter reading remains the same.

1.) Name the two electrical components in the box.

2.) State the way in which the electrical components are connected to each other.

3.) Which one of these components causes the current reading to be decreased when its connections to the cell are reversed?

4.) What is the special property possessed by this component which is responsible for the decrease in current?

Electric circuit question?

the wire must be 10/3 with ground in order to do this..... you cannot use the bare wire as a neutral....in other words there must be a black, red,white and bare wire in this wire...you must also use seperate 15 amp breakers for each hot wire red and black...you cannot use 20 amp breakers because the combined load on the neutral cannot exceed 30 amps...

Electric Circuits Question?

First, you are not transforming energy between both systems! The conservation law applies for each separate system; They are very different.

Second, in spite of the fact that you have the same power source (the same amount of energy) and bulbs, you can see in the formulas below that the current is very different in both systems by a factor of 1/4! It only means that the series circuit consumes less energy and the batteries last longer with less brightness: you can try it here too: http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/dcex3.html

You have here a simulation: http://phet.colorado.edu/simulations/sims.php?sim=Circuit_Construction_Kit_DC_Only.

More__________________________________...

Important!
1. The most bright bulb will be the one that consumes more energy per second (which means that it has more power);

2. Considerer that we have the same Voltage in both circuits and that bulbs are all equal;

In a series circuit, the power in the bulb is, P´:
P´=V´I, V' is the potential difference in one bulb and I is the electrical intensity (I is equal in both bulbs, but the voltage is the sum); because of Ohm's law, V'=R*I', but I=I', so V=R*I, and P'=R*I^2;

In a parallel circuit, the power in the bulb 1, P´ is:
P´=V*I', where V potential difference and I´ is the electrical intensity in bulb 1 (in parallel circuits the potential is the same but the current is not the same in both bulbs); well, Ohm's law says, V=R*I´, where I'=I/2 (because we considerer both bulbs equal), P'=R/4*I^2.

Conclusion: the current I in both the circuit are not equal, they are related like this: Iparallel=4*Iseries, so this shows that the power of the bulb in series (s) is and in parallel (p) is: Ps=Pp/4; which means that the brigtheness of the parallel bulbs is 4 times more!

I can only show the formulas in my site in my course of physics: physics in questions: http://www.aprenderpontofinal.com/moodle/course/view.php?id=13

More... http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/batbulb2.html#c1

Electrical circuit calculation question?

The signal LED in the secondary circuit can only with stand a limited amount of current flow before the LED will burn out. If the LED is included in the secondary circuit shown below supplied with 12V and the maximum current the LED can sustain is 0.017 Amps, what is the necessary value of the current limiting resistor RLED shown in the diagram below? Assume that voltage drop across the LED is 2.1 volts. Units are Ohms.

Electrical Circuit Question! Please Help!?

**I am going to try to draw this the best that I can :)

l------A-------B-------l
lxxxxxxxxxxxxxxl
lxxxxxxxxxxxxxxl
l-----------C-----------l
lxxxxxxxxxxxxxxl
lxxxxxxxxxxxxxxl
l----i------ -- ----------l
xxxxxx888

**I know it looks goofy, but all the dotted lines are suppose to a one continuous wire (both vertically and horizontally)....A, B, and C represent lightbulbs attached to the wires. All the 'x's are blank space, do not worry about them! I had to add those in so my picture would not be screwed up! '888' is the battery, or generator, or some form of fuel, and that is connected to the wires (I just didn't know how to draw it!) The 'i' is confusing to me, on the paper it looks as through the wire was tied together there, not 100% sure. They are connected regardless, I'm just not sure if that would change the outcome of the answers.

Please answer the following...
a) How does the brightness of each identical light bulb compare?
b) Which light bulb draws the most current?
c) What will happen if bulb A is unscrewed?
d) What will happen if bulb C is unscrewed?

Thank you VERY much!

Electrical Circuits question (Voltage divider)?

So I have this lab report for tomorrow and I'm kinda doubting my own self. Anyway...

There is a circuit with a source E = 10 V, and two resistors R1 = 10 KΩ and R2 = 2.2 ΚΩ in series. The resistors are opposite the source. We want to plug in a load RL = 100 KΩ parallel to R2 and find its wattage. In the lab, we were asked to measure the voltage across R2 and RL, which I found to be V = 1.71 V. I know it is simple, but it's hard to focus while pulling an all-nighter.

Here's what I did. I calculated the current I across RL by doing I = 1.71/100. I found this to be I = 0.71*10^-4 A. So, I multiplied this number with the voltage and found P = 0.292*10^-4 W.

Is it correct?