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Question

1) Consider circuit in figure. How much energy is absorbed by electrons from the initial state of no current (ignore thermal motion) to the state of drift velocity?
n= no of electron/volume =1029/m3,
length of circuit =10 cm,
cross-section =A=(1 mm)2



2) Electrons give up energy at the rate of RI2 per second to the thermal energy. What time scale would one associate with energy 1.78×1017 J?
n= no of electron/volume =1029/m3,
length of circuit =10 cm,
cross-section =A=(1 mm)2


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Solution

1)

Step 1: Find current in the circuit.

Formula used:V=IR

Given, voltage of the battery, V=6 V

And resistance, R=6 Ω

So current in the circuit,

I=VR=66=1 A

Step 2: Find drift velocity of the electrons.

Formula used:I=neAvd

Given,

n= no of electron/volume =1029/m3

Area of cross-section =A=(1 mm)2

So, drift velocity of the electron

vd=IneA

vd=11029×1.6×1019×106m/s

Step 3: Find energy absorbed by the electrons.

Given, length of the circuit, l=10 cm

Energy absorbed by the electron is the kinetic energy gained due to drift velocity.

E=12mv2d

Energy absorbed by all the electrons

E=12mv2d(nAl)

E=12(9.1×1031)(11029×1.6×1019×106)2(1029×106×10×102)

E=1.78×1017 J

Final answer:1.78×1017 J



2)

Step 1: Find energy given up by electron per second.

Formula used:P=I2R

Given, voltage of the battery, V=6 V

And resistance R=6 Ω

So current in the circuit, I=VR=66=1 A

So, power loss, P=I2R

P=(1)2(6)=6 W

Step 2: Find time associated.

Formula used:P=Et

Power loss is energy dissipated per second.

P=Et

t=EP=1.78×1017 J6 J/s3×1018s

Final answer:3×1018s

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