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Physics · 2023 · 5 marks
CBSE 2023 · Region 5 · Set 1 · Q31
(a) (i) Explain how free electrons in a metal at constant temperature attain an average velocity under the action of an electric field. Hence obtain…
(i)
Explain how free electrons in a metal at constant temperature attain an average velocity under the action of an electric field. Hence obtain an expression for it.
(ii)
Consider two conducting wires A and B of the same diameter but made of different materials joined in series across a battery. The number density of electrons in A is $\displaystyle 1.5$ times that in B . Find the ratio of drift velocity of electrons in wire $\displaystyle A$ to that in wire $\displaystyle B$.
(i)
A cell emf of (E) and internal resistance (r) is connected across a variable load resistance (R). Draw plots showing the variation of terminal voltage $\displaystyle V$ with (i) $\displaystyle R$ and (ii) the current (I) in the load.
(ii)
Three cells, each of emf E but internal resistances 2r, 3r and 6r are connected in parallel across a resistor $\displaystyle R$. Obtain expressions for (i) current flowing in the circuit, and
(ii)
the terminal potential difference across the equivalent cell.
Marking-scheme solution
(a)(i)
Under the effect of external field, an electron experiences a force
$$\vec{F} = -e\vec{E}$$
between collisions.
Due to this force the electron is accelerated and attains a velocity. This velocity is different for different electrons, which averaged over all electrons gives average drift velocity. This drift velocity is constant for a given temperature.
Expression of average velocity:
Under the action of an electric field electrons get accelerated with
$$a = -\frac{eE}{m}$$
Velocity of an electron at any instant of time is
$$\vec{V_i} = \vec{v_i} - \frac{e\vec{E}}{m}t_i$$
Average velocity of the electrons at time 't' is the drift velocity
$$\vec{v}_d = (\vec{V_i})_{average}$$
$$(\vec{V_i})_{average} = (\vec{v_i})_{average} - \frac{e\vec{E}}{m}(t_i)_{average}$$
But $\displaystyle (\vec{v_i})_{average} = 0$ due to randomness
$$\vec{v}_d = 0 - \frac{e\vec{E}}{m}\tau$$
$$\vec{v}_d = -\frac{e\vec{E}}{m}\tau$$
(ii)
$$v_d = \frac{I}{enA} = \left(\frac{4I}{e\pi D^2}\right) \times \frac{1}{n}$$
$$v_d \propto \frac{1}{n} \text{ for same diameter and current}$$
$$n_A = 1.5\,n_B \quad (\text{Given})$$
$$\frac{v_{dA}}{v_{dB}} = \frac{n_B}{n_A} = \frac{2}{3}$$
OR(b) (i)i.
ii.
(ii)
$$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$$
$$\frac{1}{r_{eq}} = \frac{1}{2r} + \frac{1}{3r} + \frac{1}{6r}$$
$$r_{eq} = r$$
Given cells are of equal emf (E) and connected in parallel, so
$$\frac{E_{eq}}{r_{eq}} = \frac{E}{2r} + \frac{E}{3r} + \frac{E}{6r}$$
$$E_{eq} = E$$
Current flowing $$I = \frac{E_{eq}}{R + r_{eq}}$$
$$I = \frac{E}{R + r}$$
The terminal potential difference
$$V = E_{eq} - I r_{eq}$$
$$V = E - \frac{E}{(R + r)} \times r$$
$$V = \frac{ER}{R + r}$$
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