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NCERT Exemplar · Q24

Q.A resistor R=6 ΩR = 6\ \Omega is connected across a battery of emf V=6 VV = 6\ \text{V} of negligible internal resistance, forming a single loop in which a steady current II flows.

(a) How much energy is absorbed by the conduction electrons in going from the initial state of no current (ignore thermal motion) to the steady state in which they move with the drift velocity?
(b) The electrons give up energy at the rate of RI2R I^2 per second to thermal energy. What time scale would one associate with the energy in part (a)? Take the number density of electrons n=1029 per m3n = 10^{29}\ \text{per m}^3, the length of the circuit =10 cm= 10\ \text{cm}, and the cross-sectional area A=(1 mm)2A = (1\ \text{mm})^2.
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The steady current is 1 A1\ \text{A}, corresponding to a tiny drift speed ∼6×10−5 m/s\sim 6\times10^{-5}\ \text{m/s}. Summing the drift kinetic energy of all 102210^{22} conduction electrons gives only ≈1.8×10−17 J\approx 1.8\times10^{-17}\ \text{J}. Compared with the 6 W6\ \text{W} ohmic dissipation, that energy corresponds to an astonishingly short time scale ≈3×10−18 s\approx 3\times10^{-18}\ \text{s}, far shorter than an electron's collision time — showing the drift KE is utterly negligible next to the heat continuously generated.

Given / preliminary

I=VR=66=1 A,A=(1 mm)2=10−6 m2,L=0.1 m.I = \frac{V}{R} = \frac{6}{6} = 1\ \text{A}, \qquad A = (1\ \text{mm})^2 = 10^{-6}\ \text{m}^2, \qquad L = 0.1\ \text{m}.

Drift velocity

From I=neAvdI = n e A v_d,

vd=IneA=11029×(1.6×10−19)×10−6=11.6×104=6.25×10−5 m/s.v_d = \frac{I}{n e A} = \frac{1}{10^{29}\times(1.6\times10^{-19})\times10^{-6}} = \frac{1}{1.6\times10^{4}} = 6.25\times10^{-5}\ \text{m/s}.

(a) Energy absorbed by the electrons

The number of conduction electrons in the circuit is

N=n A L=1029×10−6×0.1=1022.N = n\,A\,L = 10^{29}\times10^{-6}\times0.1 = 10^{22}.

Each acquires drift kinetic energy 12mvd2\tfrac12 m v_d^2 (with m=9.1×10−31 kgm = 9.1\times10^{-31}\ \text{kg}), so the total energy absorbed is

KE=12Nmvd2=12(1022)(9.1×10−31)(6.25×10−5)2.KE = \tfrac12 N m v_d^2 = \tfrac12 (10^{22})(9.1\times10^{-31})(6.25\times10^{-5})^2.

(6.25×10−5)2=3.906×10−9,KE=12(1022)(9.1×10−31)(3.906×10−9)≈1.8×10−17 J.(6.25\times10^{-5})^2 = 3.906\times10^{-9}, \quad KE = \tfrac12 (10^{22})(9.1\times10^{-31})(3.906\times10^{-9}) \approx 1.8\times10^{-17}\ \text{J}.

(b) Associated time scale …

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