  Question

Two  metallic wires $$P_{1}$$  and $$P_{2}$$  of the same material and  same   length but  different cross -  sectional    areas  $$A_{1}$$  and $$A_{2}$$  are joined together  and  then connected to a source of emf.   find the ratio of the drift.  Velocities of free electrons in  the wires $$P_{1}$$  and $$P_{2}$$  if the wires are connected (i)  in series,  and  (ii)  in parallel.

Solution

We know that,$$I=neA\,v_d \Rightarrow v_d=\dfrac{1}{neA}$$Let $$R_1$$ and $$R_2$$ be resistance of $$P_1$$ and $$P_2$$ and $$A_1$$ and $$A_2$$ are their cross sectional areas respectively.$$\therefore R_1=p\dfrac{1}{A_1}$$ and $$R_2=p\dfrac{1}{A_2}$$When connected in seried $$\therefore \dfrac{v_{d_1}}{v_{d_2}}=\dfrac{\dfrac{c}{(\dfrac{p^l}{A_1}+\dfrac{p^l}{A_2})naA_1}}{\dfrac{c}{(\dfrac{pl}{A_1}+\dfrac{p^l}{A_2})neA_2}}=\dfrac{A_2}{A_1}$$When,connected in parallel,$$\dfrac{v_{d_1}}{v_{d_2}}=\dfrac{\dfrac{\dfrac{\varepsilon}{pl}.\dfrac{1}{neA_1}}{A_1}}{\dfrac{\varepsilon}{\dfrac{pl}{A_2}}.\dfrac{1}{neA_2}}=1$$ Physics

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