Two metallic wires A and B made from same substance have equal le… - Vidyadip

Question

Two metallic wires A and B made from same substance have equal lengths and their cross-sections are in the ratio $1: 2$. They are joined in (i) series (ii) parallel. In both the above combinations, compare the drift velocities of electrons in the two wires.

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Answer

$A _1: A _2=1: 2=l_2=1$
(i) On joining in series, the electric current is same.
$I _{ A }= I _{ B } \Rightarrow n e A_1\left(v_d\right)_1=n e A_2\left(v_d\right)_2$
or $\frac{\left(v_d\right)_1}{\left(v_d\right)_2}=\frac{ A _2}{A_1}=\frac{2}{1}=2: 1$
(ii) In parallel combination, potential difference remains the same
Hence, $V_A=V_B \Rightarrow I_A \times R_A=I_B \times R_B$
$\Rightarrow n e A_1\left(v_d\right)_1 \times \rho \frac{1}{A_1}=n e A_2\left(v_d\right)_2 \times \rho \frac{1}{A_2}$
$\Rightarrow \quad \frac{\left(v_d\right)_1}{\left(v_d\right)_2}=1 \Rightarrow\left(v_d\right)_1:\left(v_d\right)_2=1 ; 1$ Ans.

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