Charge, Current & Power Homework Statement

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SUMMARY

The discussion focuses on calculations related to a smartphone powered by a 3.82V battery delivering 1.81A of current. The amount of charge delivered by the battery is calculated to be 6516C, while the charge before the phone loses power is 6385.68C. The relationship between charge, voltage, and energy consumption is established, confirming that energy consumed is equal to QV. Additionally, the discussion explains how intrinsic semiconductors increase conductivity with temperature and the role of holes in current flow.

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  • Familiarity with semiconductor physics, particularly intrinsic and extrinsic semiconductors.
  • Knowledge of power calculations in electrical circuits.
  • Ability to perform unit conversions and basic algebraic manipulations.
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BoanviaFx
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Homework Statement


Can someone please confirm if I answered correctly?

Question: According to rumours the next generation smart phone will be powered by a 3.82V battery that can deliver 1.81A of current for one hour. The phone automatically turns off when the battery has less than 2% of charge remaining.

i) Calculate the amount of charge that can be delivered by the battery.
ii) How much charge has passed before the phone loses power?
iii) Show that if a charge Q is transferred across a p.d V, the energy consumed is equal to QV.
iv) In an idle state the phone uses a constant 70mW of power. How many days will pass before the phone loses power completely?
v) The majority of components inside a smart phone are made up of semiconducting material. Briefly explain how the electrical conductivity of an intrinsic semiconductor increases as the temperature rises.
vi) Explain the origin of holes in intrinsic semiconducting materials and explain the process by which holes participate in current flow.

Homework Equations

The Attempt at a Solution


i) Solution: I=Q/T Q=IT Q=(1.81*60*60)= 6516C

ii) 100%-2% = 98%
6516*0.98= 6385.68C

iii) P=IV = (1.81*3.82) = 6.9142W
Calculating Charge at 1 second:
I=Q/T Q=IT (1.81*1) = 1.81C
Therefore P=QV = (1.81*3.82) = 6.9142W

iv) P=IV I=P/V I=70mW/3.82V = 0.0183A
Therefore charge every second = 0.0183C
Total time = 6516C/0.0183C = 356065.6s
356065.6/3600 = 98.9 hours.

v) Upon temperature rise, valence electrons gain energy and break free becoming free electrons. The more free electrons the greater the conductivity. The semiconductor is known as negative temperature coefficient. Temperature rises and resistance decreases. Since resistance decreases so does the resistivity. Conductivity is the reciprocal of resistivity therefore conductivity will increase.

vi) When the atom has absence of electrons it means it has a hole. Therefore the atom would lack any free electrons, this would will result in no current flow making it an insulator.
 
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(iii) the energy consumed is E=VIt assuming that the current is constant. By replacing I=Q/t you have E=QV.

(vi) I think you understood it wrong, holes behave as current carriers like the free electrons. A P-type semiconductor has holes as the current carriers while a N-type has free electrons .
 
Oh my bad, yes you are right. Holes travel the opposite direction from a higher potential to a lower potential. They too help with conductivity. Delta thank you for correcting my mistakes this helps me a lot. :)
 

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