I have done problems 1, 2, and 3, if someone could help me with 4, 5 and 6 that

Question

I have done problems 1, 2, and 3, if someone could help me with 4, 5 and 6 that would be great!

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 40 W light bulbs, a 100 W light bulb and a 1400 W hairdryer together with four switches as shown. The voltage supply is from a plug in the wall (120 V). (The plug supplies AC current, but we will assume for simplicity it is DC as shown.) Important: 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 a bulb, a different amount of power will be used. (But, regardless of the voltage applied, the resistance of any object remains constant. So, we can think of each appliance as a given resistor sitting in the circuit.) 120V Switches S1, S2, Ss, and Ss can be open (no current will flow) or closed (current flows through the switch freely) Resistors R, and R are the 40 W bulbs, R2 is the 1400 W hairdryer and Ra is the 100 W light bulb. Switch S4 is open for now. The resistor r is 0.1 . It represents a possible short-circuit that develops if switch Sa is closed. For each problem, show all algebraic steps.

Explanation / Answer

resistance values:

resistance of 40 W light bulbs=voltage^2/power=120^2/40=360 ohms

==>R1=R3=360 ohms

resistance of 100 W light bulb=120^2/100=144 ohms

==>R4=144 ohms

resistance of 1400 W hair dryer=120^2/1400=10.2857 ohms

R2=10.2857 ohms

Q4.

given that S1 and S2 are closed and other switches are open

then R1 and R2 are connected in parallel across 120 volts supply

as each reistor has supply voltage of 120 volts, they are operating normally

hence total power consumed by the two appliences=(120^2/R1)+(120^2/R2)

=40+1400=1440 W

current supplied by the 120 volt source=power/voltage

=1440/120=12 A

Q5.as S4 is closed, 0.1 ohm is conencted across 120 volt supply

power dissipated across it=120^2/0.1=144000 W

total power consumed=144000+1440=145440 W

current supplied by the source=power/voltage=1212 A

as the current and power values are very high, the supply wont be able to sustain these high demand and a spark will occur due to thermal overload.

almost all of the power is being disispated in the 0.1 ohm resistance

Q6.
power dissipated in the chewed cord is large.

this much power generated will quickly raise the temperature and a thermal overload and spark will occur.

a fuse is provided just next to the power supply so that

it will be blown once the current demand value increases beyond a particular maximum safe value.

no,the analysis does not depend upon location of short circuit .

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