Ncert Exercises & Solutions

(a) Consider the 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?

(b) Electrons give up energy at the rate of ${RI}^{2}$ per second to the thermal energy. What time scale would one associate with energy in problem (a)?
n = number of electron/volume = $10^{29} / m^{3}$ , Length of circuit = 10 cm, cross-section: A = ${(1 mm)}^{2}$

Hint: Apply Ohm's law.

Step 1: By ohm's law, current I is given by;

I = 6 V / 6 Ω = 1 A

But, I = ne {Av}_{d}

or v_{d} = \frac{i}{neA}

On substituting the values;
For n = number of electron/volume =

10^{29} m^{3}

length ot circuit 1O cm, cross-section, A =

{(1 mm)}^{2}

\begin{array}{l} v_{d} = \frac{1}{10^{29} \times 1 .6 \times 10^{- 19} \times 10^{- 6}} \\ = \frac{1}{1 .6} \times 10^{- 4} m / s \end{array}

Step 2: Therefore, the energy absorbed in the form of KE is given by:

\begin{array}{l} KE = \frac{1}{2} m_{e} v_{d}^{2} \times nAI \\ = \frac{1}{2} \times 9 .1 \times 10^{- 31} \times \frac{1}{2 .56} \times 10^{29} \times 10^{- 8} \times 10^{- 6} \times 10^{- 1} \\ = 1.77 \times 10^{- 17} J \end{array}

Step 3:

(b) Power loss is given by;

P = I^{2} R = 6 \times 1^{2} = 6 W = 6 J / s

\begin{array}{ll} {Since}_{,} & P = \frac{E}{t} \\ Therefore, & E = P \times t \\ or & t = \frac{E}{P} = \frac{1.77 \times 10^{- 17}}{6} = 0.3 \times 10^{- 17} s \end{array}

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