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 $={10}^{29}/m^3$,
length of circuit $=10cm$,
cross-section $=A=(1mm{)}^2$



2) Electrons give up energy at the rate of $R{I}^2$ per second to the thermal energy. What time scale would one associate with energy $1.78\times {10}^{-17}J$?
$n=$ no of electron/volume $={10}^{29}/m^3$,
length of circuit $=10cm$,
cross-section $=A=(1mm{)}^2$

Answer 1

1)

$\text{Step 1: Find current in the circuit}.$

$\text{Formula used}:V=IR$

Given, voltage of the battery, $V=6V$

And resistance, $R=6\Omega$

So current in the circuit,

$I=\frac{V}{R}=\frac{6}{6}=1A$

$\text{Step 2: Find drift velocity of the electrons}.$

$\text{Formula used}:I=neA{v}_d$

Given,

$n=$ no of electron/volume $={10}^{29}/m^3$

Area of cross-section $=A=(1mm{)}^2$

So, drift velocity of the electron

$v_d=\frac{I}{neA}$

$v_d=\frac{1}{{10}^{29}\times 1.6\times {10}^{-19}\times {10}^{-6}}m/s$

$\text{Step 3: Find energy absorbed by the electrons}.$

Given, length of the circuit, $l=10cm$

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

$E=\frac{1}{2}m{v}_d^2$

Energy absorbed by all the electrons

$E=\frac{1}{2}m{v}_d^2\left(nAl\right)$

$E=\frac{1}{2}(9.1\times {10}^{-31}){\left(\frac{1}{{10}^{29}\times 1.6\times {10}^{-19}\times {10}^{-6}}\right)}^2({10}^{29}\times {10}^{-6}\times 10\times {10}^{-2})$

$E=1.78\times {10}^{-17}J$

$\text{Final answer}:1.78\times {10}^{-17}J$



2)

$\text{Step 1: Find energy given up by electron per second.}$

$\text{Formula used}:P=I^{2}R$

Given, voltage of the battery, $V=6V$

And resistance $R=6\Omega$

So current in the circuit, $I=\frac{V}{R}=\frac{6}{6}=1A$

So, power loss, $P=I^{2}R$

$P=\left(1{)}^2\right(6)=6W$

$\text{Step 2: Find time associated}.$

$\text{Formula used}:P=\frac{E}{t}$

Power loss is energy dissipated per second.

$P=\frac{E}{t}$

$t=\frac{E}{P}=\frac{1.78\times {10}^{-17}J}{6J/s}\approx 3\times {10}^{-18}s$

$\text{Final answer}:3\times {10}^{-18}s$