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प्रश्न
- 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?
- 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
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उत्तर
a. Relation between current and drift velocity is given by I = ne Avd, where vd 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 Avd
or vd = `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.
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संबंधित प्रश्न
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