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

Answer 1

a. Relation between current and drift velocity is given by I = ne Av_d, where v_d is the drift speed of electrons and n is the number density of electrons.

According to the Ohm's law current in the circuit

I = V/R

I = 6 V/6 Ω = 1 A

But, I = ne Av_d

or v_d = `I/("ne A")`

On substituting the values,

For, n = number of electrons/volume = 10²⁹/m³

Length of circuit = 10 cm, cross-section = A = (1 mm)²

`v_d = 1/(10^29 xx 16 xx 10^-19 xx 10^-6)`

= `1/1.6 xx 10^-4` m/s

Therefore, the energy absorbed in the form of KE is given by

Total KE = KE of 1 electron × no. of electrons

KE = `1/2 m_e v_d^2 xx n Al`

= `1/2 xx 9.1 xx 10^31 xx 1/2.56 xx 10^-8 xx 10^29 xx 10^-6 xx 10^-1`

= `2 xx 10^-17 J`

b. Ohmic loss (power loss) is `P = I^2R = 6 xx 1^2 = 6 W = 6 J/s`

Since, the energy dissipated per unit time is the power dissipated.

So, `P = E/t`

Therefore, `E = P xx t`

or `t = E/P = (2 xx 10^17)/6 = 10^17 s` 

Important point: The energy dissipated per unit time is the power dissipated `P = (ΔW)/(Δt)`

The power across a resistor is P = I²R.