In your house or apartment, each appliance is wired in parallel. However, suppos
ID: 1355677 • Letter: I
Question
In your house or apartment, each appliance is wired in parallel. However, suppose from a single outlet you unwisely decided to do some "adventurous" wiring. You wire two 60W light bulbs, a 100W light bulb and a 1200 W hairdryer together with four switches as shown. The voltage supply is from a plug in the wall (120 V). (The plus supplies AC current, but we will assume for simplicity it is DC as shown).
Note that the wattage value listed on the bulb package assumes that 120 volts will be applied across the bulb. If a different voltage is applied to the bulb (or hairdryer), a different amount of power will be used. Regardless of the voltage applied, we shall assume for this homework that the resistance of each remains constant (note that this isn’t exactly true, as you will find in the lab this week). So, for this homework we can think of each appliance as a resistor sitting in the circuit.
HERE IS THE URL FOR THE DIAGRAM/IMAGE OF THE CIRCUIT:
http://www.colorado.edu/physics/phys2020/phys2020_fa15/Written%20HW/Physics%202020%20HW%205.pdf
Switches S1, S2, S3, and S4 can be open (no current will flow) or independently closed (current flows through that branch of the switch freely). Resistors R1 and R3 are the 60 W bulbs, R2 is the 1200 W hairdryer and R4 is the 100 W light bulb. Switch S4 is open for now. The resistor r is 1 Ohm. It represents a possible short-circuit that develops if switch S4 is closed.
FOR EACH PROBLEM, SHOW ALL ALGEBRAIC STEPS!
1). When switch S1 is closed and the others are all open, how much power is dissipated in bulb R1? This will be the total power supplied by the wall socket. When switch S2 is now closed in addition to S1, how much power does the wall socket now supply? What does this tell you about how powers add for parallel resistors? Does it matter whether the switches are above or below the resistor in the schematic (Why or why not)?
2). Determine the resistances of R1 through R4. If you have already done a portion of that to answer question 1 above (you didn’t need to), you can copy those numbers down here.
3. Suppose only switch S3 is closed. Determine the total current and power supplied by the wall socket. Be careful about this one, which is trickier than the above case. Determine the power dissipated in each bulb individually (R3 and R4). Do the bulbs light up properly? Is the 100 W bulb or the 60 W bulb brighter? Why? Would it matter if you switched the positions of R3 and R4 in the schematic Why or why not?
4. Suppose switches S1 and S2 are closed as in the previous part. Also suppose your cat chews on the power cord the wires in the cord touch, the effect is the same as closing switch S4 so that the 1 Ohm short-circuit resistance becomes part of the circuit. (The cat is quick and gets away safely.) Determine the current and power supplied by the voltage supply. Where in the circuit is almost all of the power dissipated?
5. Does it seem like the power dissipated in the chewed cord is large or small? Explain why this situation is very hazardous (this is why you have circuit breakers in your house!). Would your analysis change if the short-circuit was on the left side rather than the right side of the schematic?
Explanation / Answer
part 1:
resistance=voltage^2/power rating
resistances of different appliances are:
R1=120^2/60=240 ohms
R2=120^2/1200=12 ohms
R3=120^2/60=240 ohms
R4=120^2/100=144 ohms
when switch S1 is closed and others are open:
then only R1 will be conencted to the voltage ource.
then power dissipated in bulb R1=voltage^2/R1=120^2/240=60 W
when switch S2 is also closed:
voltage across R1=120 volts
then power supplied to R1=120^2/R1=120^2/240=60 W
voltage across R2=120 volts
power supplied to R2=120^2/R2=1200 W
hence net power supplied=60+1200=1260 W
hence for parallel resistors, powers are added .
it does not matter where the switches are because unless the switch is closed and the circuit is completed,
there will no flow of current.
Q2.
R1=120^2/60=240 ohms
R2=120^2/1200=12 ohms
R3=120^2/60=240 ohms
R4=120^2/100=144 ohms
Q3.if only switch S3 is closed, then both R3 and R4 are connected to the voltage source.
then net resistance=R3+R4=384 ohms
then current supplied=120/384=0.3125 A
power dissipated in R3=current^2*R3=23.4375 W
power dissipated in R4=current^2*R4=14.0625 W
no,the bulbs dont light up properly as the dont get their rated power.
as bulb 3 gets more power (closer to its rated value than in case of R4), it will be brighter.
position of switch does not matter as without the circuit being closed, there wont be any current.
hence no power is provided from the source.
Q4. switch S1,,S2 and S4 are closed.
then R1,R2 and r are connected to the source.
then current supplied to R1=120/R1=0.5 A
power supplied to R1=120^2/R1=60 W
current supplied to R2=120/R2=10 A
power supplied to R2=120^2/R2=1200 W
current supplied to r=120/r=120 A
power supplied to r=120^2/r=14400 W
hence total current supplied=0.5+10+120=130.5 A
total power supplied=60+1200+14400=15660 W
almost all the power is dissipated in r (91.95% of total power)
Q5.power dissipated is very large.
as it draws current of 120 A, which is very large current, it will create sparking and it can damage the conductors permanently due to heating.
it does not matter where the resistance r is in the schematics.