Source: High school physics (Chinese)
Problem
Two copper wires with different diameters are connected in series and carry a constant current. The ratio of their diameters is $d_1 : d_2 = 3 : 2$, and the ratio of their lengths is $l_1 : l_2 = 1 : 2$.
- Find the ratio of current densities.
- Find the ratio of electric field strengths inside the wires.
- Find the ratio of free-electron drift speeds.
- Find the ratio of voltages across the two wires.
- Find the ratio of electric energy consumed in one hour.
(1) $j_1 : j_2 = 4 : 9$; \quad (2) $E_1 : E_2 = 4 : 9$; \quad (3) $v_1 : v_2 = 4 : 9$; \quad (4) $U_1 : U_2 = 2 : 9$; \quad (5) $W_1 : W_2 = 2 : 9$.
Series connection means the same current $I$ flows through both wires. The cross-sectional area scales as the square of the diameter, so:
$$\frac{A_1}{A_2} = \left(\frac{d_1}{d_2}\right)^2 = \frac{9}{4}$$Both wires are copper, so the resistivity $\rho$ and free-electron density $n$ are identical.
(1) Current density $j = I/A$:
$$\frac{j_1}{j_2} = \frac{A_2}{A_1} = \frac{4}{9}$$(2) Electric field $E = \rho j$ (Ohm's law in microscopic form), with the same $\rho$:
$$\frac{E_1}{E_2} = \frac{j_1}{j_2} = \frac{4}{9}$$(3) Drift speed $v = j/(n e)$, with the same $n$ and $e$:
$$\frac{v_1}{v_2} = \frac{j_1}{j_2} = \frac{4}{9}$$(4) Voltage $U = E\, l$:
$$\frac{U_1}{U_2} = \frac{E_1}{E_2} \cdot \frac{l_1}{l_2} = \frac{4}{9} \cdot \frac{1}{2} = \frac{2}{9}$$(5) Energy $W = U I t$ with the same $I$ and $t$:
$$\frac{W_1}{W_2} = \frac{U_1}{U_2} = \frac{2}{9}$$