Two metallic wires of the same length but cross-sectional area is in the ratio 1:2.They are (i)in series and (ii) in parallel.Compute the drift velocities of electrons in the two wires in both the cases.

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Solution

$D r i f t v e l o c i t y i s g i v e n b y : V_{d} = \frac{I}{A n e} w h e r e A i s t h e a r e a o f t h e c o n d u c t o r . w h e n t w o c o n d u c t o r s a r e p a r a l l e l t h e n n e t a r e a o f t h e c o n d u c t o r i s s a m e h e n c e d r f i t v e l o c i t y o f t h e e l e c t r o n s i s s a m e t h r o u g h b o t h t h e c o n d u c t o r . n e t a r e a = A, = A_{1} + A_{2} = 3 A V_{d 1} = \frac{I}{A' n e} = V_{d 1} = \frac{I}{3 A n e} V_{d 2} = \frac{I}{A' n e} = \frac{I}{3 A n e} W h e n t h e y a r e i n s e r i e s, A r e a o f o n e c o n d u c t o r i s t w i c e t o t h e a n o t h e r t h e r e f o r e d r i f t v e l o c i t y o f t h e c o n d u c t o r h a v i n g l e s s a r e a i s m o r e t h a n t h e o t h e r . A_{1} = A A_{2} = 2 A V_{d 1} = \frac{I}{A_{1} n e} = \frac{I}{A n e} V_{d 2} = \frac{I}{A_{2} n e} = \frac{I}{2 A n e} \frac{V_{d 1}}{V_{d 2}} = 1 / 2$

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