Question

Two metallic wires of the same material and same length but different cross rational areas are joined together ( in series) and ( in parallel ) ,to a source of amf. in which of the two wires will the drift velocity of electron be more in each of the two cases and why?

Answer 1

Answer on Question #46854 – Physics – Solid State Physics

Two metallic wires of the same material and same length but different cross rational areas are joined together (in series) and (in parallel), to a source of amf. in which of the two wires will the drift velocity of electron be more in each of the two cases and why?

Solution:

The formula for evaluating the drift velocity of electrons in a material of constant cross-sectional area is given by (A – area of cross – section of the material, I – current flowing through the material, q – the charge of the electron, n is the charge-carrier density):

$$v = \frac {I}{n A q}$$

First case: series connection

In a series circuit, the current through each of the components is the same $(I_1 = I_2)$, thus drift velocity is more in wire with smaller area of cross-section:

$$v_1 = \frac {I}{n A_1 q}$$ $$v_2 = \frac {I}{n A_2 q}$$ $$A_2 > A_1 \Rightarrow \frac {I}{n A_2 q} < \frac {I}{n A_1 q}$$ $$v_2 < v_1$$

First case: parallel connection

In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component (L – length of the wire, $\rho$ – electrical resistivity).

$$I_1 = \frac {U}{R} = \frac {U}{\frac {\rho L}{A_1}} = \frac {U A_1}{\rho L}$$ $$I_2 = \frac {U}{R} = \frac {U}{\frac {\rho L}{A_2}} = \frac {U A_2}{\rho L}$$ $$v_1 = \frac {I_1}{n A_1 q} = \frac {1}{n A_1 q} \frac {U A_1}{\rho L} = \frac {U}{n q L}$$ $$v_2 = \frac {I}{n A_2 q} = \frac {1}{n A_2 q} \frac {U A_2}{\rho L} = \frac {U}{n q L}$$

According to initial condition of the task, $L_1 = L_2 = L$, $\rho_1 = \rho_2 = \rho$ (wires of the same material and same length), thus, drift velocities will be the same in both wires.

$$v_1 = v_2 = \frac {U}{n q L}$$

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