CBSE 2024 · Region 3 · Set 1 · Q17 · 2 marks
What is meant by 'relaxation time' of free electrons in a conductor ? Show that the resistance of a conductor can be expressed by $\displaystyle \mathrm{R}=\frac{\mathrm{m} l}{\mathrm{ne}^{2} \tau \mathrm{~A}}$, where symbols have their usual meanings.Draw the circuit diagram of a Wheatstone bridge. Obtain the condition when no current flows through the galvanometer in it.
What is meant by 'relaxation time' of free electrons in a conductor ? Show that the resistance of a conductor can be expressed by $\displaystyle \mathrm{R}=\frac{\mathrm{m} l}{\mathrm{ne}^{2} \tau \mathrm{~A}}$, where symbols have their usual meanings.
Draw the circuit diagram of a Wheatstone bridge. Obtain the condition when no current flows through the galvanometer in it.
Marking-scheme solution
Average time between two successive collisions of electron in presence of electric field.
Drift velocity of an electron
\[\upsilon_d = \frac{eE}{m}\tau \qquad ---(i) \]
Current flowing through a conductor of length $\displaystyle l$ and area of cross section A
\[I = neA\upsilon_d \qquad ---(ii) \]
\[I = \frac{ne^2 AE\tau}{m} = \frac{ne^2 A\tau V}{ml} \]
\[R = \frac{V}{I} = \frac{ml}{ne^2 \tau A} \]
By applying Kirchoff's loop rule to closed loops ADBA and CBDC
$\displaystyle -I_1R_1 + 0 + I_2R_2 = 0$ -----(i) $\displaystyle [I_g = 0]$
$\displaystyle I_2R_4 + 0 - I_1R_3 = 0$ -----(ii)
From eq (i)-
\[\frac{I_1}{I_2} = \frac{R_2}{R_1} \]
From eq (ii)-
\[\frac{I_1}{I_2} = \frac{R_4}{R_3} \]
Hence,
\[\frac{R_2}{R_1} = \frac{R_4}{R_3} \]
Current ElectricityDrift of Electrons and the Origin of ResistivityApplyvery_short_answermedium
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