A physics student has a single-occupancy dorm room. The student has a small refr

Question

A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A and a voltage of 110 V, a lamp that contains a 100-W bulb, an overhead light with a 60-W bulb, and various other small devices adding up to 3.00 W. (a) Assuming the power plant that supplies 110 V electricity to the dorm is 10 km away and the two aluminium transmission cables (take rho_A1 = rho_AI, 0 = 2.82 times 10^8 ohm middot m) use wire with h diameter of 8.25 mm (0 gauge), estimate the percentage of the total power supplied by the power company that is lost in the transmission, (b) What would be the result if the power company delivered the electric power at 110 kV?

Explanation / Answer

part a:

power consumed by refrigerator=voltage*current=330 W

total power consumed=330+100+60+3=493 W

total current consumed=power /voltage=4.4818 A

resistance of each wire=resistivity*length/area

=2.82*10^(-8)*10*1000/(pi*(8.25*0.001/2)^2)=5.2754 ohms

as the wires are connected in parallel,

equivalent resistance=5.2754/2=2.6377 ohms

then power lost in transmission=current^2*resistance

=4.4818^2*2.6377=52.982 W

then percentage power lost=

=power lost/total power generated

=power lost/(power consumed+power lost)

=52.982/(52.982+493)

=9.704%

part b:

if power is delivered at 110 kV,

then total power consumed=110*10^3*3+100+60+3=330163 W

current=330163/110000=3.0015 A

total power lost=3.0015^2*2.6377=23.763 W

percentage power lost=23.763*100/(23.763+330163)

=7.1968*10^(-3) %

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.