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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,
$$\dfrac{v_{d_1}}{v_{d_2}}=\dfrac{\dfrac{\dfrac{\varepsilon}{pl}.\dfrac{1}{neA_1}}{A_1}}{\dfrac{\varepsilon}{\dfrac{pl}{A_2}}.\dfrac{1}{neA_2}}=1$$
1663474_1606607_ans_bf0d8353585340bfa9fd503626aa1ac2.png

Physics

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