Chapter 3: Current Electricity

(a) Drift speed.

The current is related to the drift speed by $$I = n e A v_d$$, where $$n$$ is the number density of conduction electrons, $$e$$ the electronic charge, $$A$$ the cross-sectional area and $$v_d$$ the drift speed. Hence $$v_d = \dfrac{I}{n e A}$$.

Step 1 — find the electron number density $$n$$. The number of copper atoms per unit volume is $$\dfrac{\rho_m N_A}{M}$$, and since each atom contributes one conduction electron, $$n$$ equals this quantity. With density $$\rho_m = 9.0 \times 10^{3} \, \mathrm{kg/m^3}$$, Avogadro number $$N_A = 6.0 \times 10^{23} \, \mathrm{mol^{-1}}$$ and atomic mass $$M = 63.5 \, \mathrm{u} = 63.5 \times 10^{-3} \, \mathrm{kg/mol}$$:

$$n = \dfrac{(9.0 \times 10^{3})(6.0 \times 10^{23})}{63.5 \times 10^{-3}} \approx 8.5 \times 10^{28} \, \mathrm{m^{-3}}$$

Step 2 — find $$v_d$$. Substituting $$I = 1.5 \, \mathrm{A}$$, $$e = 1.6 \times 10^{-19} \, \mathrm{C}$$ and $$A = 1.0 \times 10^{-7} \, \mathrm{m^2}$$:

$$v_d = \dfrac{1.5}{(8.5 \times 10^{28})(1.6 \times 10^{-19})(1.0 \times 10^{-7})} = \dfrac{1.5}{1.36 \times 10^{3}}$$

$$v_d \approx 1.1 \times 10^{-3} \, \mathrm{m\,s^{-1}} = 1.1 \, \mathrm{mm\,s^{-1}}$$

(b) Comparisons.

(i) With the thermal speed of the atoms. At an ordinary temperature ($$T \approx 300 \, \mathrm{K}$$) the random thermal speed of a copper atom is of the order $$v_{th} \sim \sqrt{\dfrac{3 k_B T}{M_{atom}}} \sim 10^{2} \, \mathrm{m\,s^{-1}}$$. The drift speed is therefore smaller than the thermal speed by a factor of about $$10^{-5}$$.

(ii) With the speed of propagation of the field. When the circuit is closed the electric field is set up throughout the conductor with a speed close to that of light, $$\sim 3 \times 10^{8} \, \mathrm{m\,s^{-1}}$$. The drift speed is smaller than this by a factor of about $$10^{-11}$$.

When a negligibly small current flows, the element is at room temperature $$T_1 = 27.0 \, °\mathrm{C}$$ and its resistance is $$R_1 = 75.3 \, \Omega$$.

When the toaster is connected to the supply, at the steady temperature $$T_2$$ the resistance is obtained from Ohm's law:

$$R_2 = \dfrac{V}{I} = \dfrac{230 \, \mathrm{V}}{2.68 \, \mathrm{A}} \approx 85.8 \, \Omega$$

The resistance varies with temperature as $$R_2 = R_1\left[1 + \alpha (T_2 - T_1)\right]$$, so

$$T_2 - T_1 = \dfrac{R_2 - R_1}{R_1 \, \alpha}$$

$$T_2 - T_1 = \dfrac{85.8 - 75.3}{75.3 \times 1.70 \times 10^{-4}} = \dfrac{10.5}{1.28 \times 10^{-2}} \approx 820 \, °\mathrm{C}$$

$$T_2 = 27 + 820 = 847 \, °\mathrm{C}$$

For a platinum resistance thermometer the resistance is taken to vary linearly with temperature. The temperature $$t$$ corresponding to a resistance $$R_t$$ is found by linear interpolation between the ice point ($$0 \, °\mathrm{C}$$) and the steam point ($$100 \, °\mathrm{C}$$):

$$t = \dfrac{R_t - R_0}{R_{100} - R_0} \times 100 \, °\mathrm{C}$$

Here $$R_0 = 5 \, \Omega$$ (ice point), $$R_{100} = 5.23 \, \Omega$$ (steam point) and $$R_t = 5.795 \, \Omega$$ (hot bath). Substituting:

$$t = \dfrac{5.795 - 5}{5.23 - 5} \times 100 = \dfrac{0.795}{0.23} \times 100$$

$$t = 3.4565 \times 100 \approx 345.65 \, °\mathrm{C}$$

The network of 12 equal resistors cannot be reduced by simple series–parallel rules, but the symmetry of the cube about the diagonal can be exploited.

Using symmetry and the junction rule. Let a total current $$3I$$ enter at corner A and leave at the diagonally opposite corner C'. The three edges meeting at A (namely AA', AD and AB) are symmetrically placed, so each must carry the same current $$I$$.

At the next corners (A', D and B) each incoming current $$I$$ splits equally into the two outgoing edges, so those edges each carry $$I/2$$. By the same symmetry, at the three corners adjacent to C' (D', B' and C) two currents of $$I/2$$ recombine, so each of the three edges leading into C' again carries $$I$$. This fixes the current in every one of the 12 edges.

Applying the loop rule. Take the closed loop A → B → C → C' (each edge has resistance $$R$$) together with the battery of emf $$\varepsilon$$. Going around:

$$\varepsilon = I R + \dfrac{I}{2} R + I R = \dfrac{5}{2} I R$$

The equivalent resistance is the emf divided by the total current $$3I$$ drawn from the battery:

$$R_{eq} = \dfrac{\varepsilon}{3I} = \dfrac{(5/2) I R}{3 I} = \dfrac{5}{6} R$$

With $$R = 1 \, \Omega$$:

$$R_{eq} = \dfrac{5}{6} \, \Omega \approx 0.83 \, \Omega$$

Currents along the edges. The total current drawn from the $$10 \, \mathrm{V}$$ battery is

$$3I = \dfrac{\varepsilon}{R_{eq}} = \dfrac{10}{5/6} = 12 \, \mathrm{A} \quad\Rightarrow\quad I = 4 \, \mathrm{A}$$

Hence the three edges meeting at A and the three edges meeting at C' each carry $$I = 4 \, \mathrm{A}$$, while each of the remaining six (middle) edges carries $$I/2 = 2 \, \mathrm{A}$$.

Assign a current to each branch and reduce the number of unknowns by applying the junction rule at every junction as the currents are labelled. As in Fig. 3.17, take $$I_1$$, $$I_2$$ and $$I_3$$ as the three independent currents; the current in every branch is then written in terms of them.

Loop ADCA. Applying Kirchhoff's loop rule:

$$10 - 4(I_1 - I_2) + 2(I_2 + I_3 - I_1) - I_1 = 0$$

$$\Rightarrow \; 7 I_1 - 6 I_2 - 2 I_3 = 10 \quad\text{...(1)}$$

Loop ABCA.

$$10 - 4 I_2 - 2(I_2 + I_3) - I_1 = 0$$

$$\Rightarrow \; I_1 + 6 I_2 + 2 I_3 = 10 \quad\text{...(2)}$$

Loop BCDEB.

$$5 - 2(I_2 + I_3) - 2(I_2 + I_3 - I_1) = 0$$

$$\Rightarrow \; 2 I_1 - 4 I_2 - 4 I_3 = -5 \quad\text{...(3)}$$

Solving the three equations. Adding (1) and (2):

$$8 I_1 = 20 \quad\Rightarrow\quad I_1 = 2.5 \, \mathrm{A}$$

Put $$I_1 = 2.5$$ in (2): $$2.5 + 6 I_2 + 2 I_3 = 10$$, i.e. $$3 I_2 + I_3 = 3.75$$.
Put $$I_1 = 2.5$$ in (3): $$5 - 4 I_2 - 4 I_3 = -5$$, i.e. $$I_2 + I_3 = 2.5$$.

Subtracting the second from the first: $$2 I_2 = 1.25$$, so

$$I_2 = \dfrac{5}{8} \, \mathrm{A} = 0.625 \, \mathrm{A}, \qquad I_3 = 2.5 - 0.625 = \dfrac{15}{8} \, \mathrm{A} = 1.875 \, \mathrm{A}$$

Current in each branch.

It may be verified that these values satisfy Kirchhoff's loop rule for every other closed loop of the network as well.

Let $$I_1$$ be the current entering at A through arm AB and $$I_2$$ the current entering at A through arm AD. Let $$I_g$$ be the current through the galvanometer, taken from B to D. By the junction rule the current in BC is $$I_1 - I_g$$ and the current in DC is $$I_2 + I_g$$.

Mesh BADB. Applying the loop rule around B → A → D → B:

$$100 I_1 + 15 I_g - 60 I_2 = 0$$

Dividing by 5: $$\;\; 20 I_1 + 3 I_g - 12 I_2 = 0 \quad\text{...(a)}$$

Mesh BCDB. Around B → C → D → B:

$$10(I_1 - I_g) - 5(I_2 + I_g) - 15 I_g = 0$$

$$10 I_1 - 30 I_g - 5 I_2 = 0 \;\Rightarrow\; 2 I_1 - 6 I_g - I_2 = 0 \quad\text{...(b)}$$

Mesh ADCEA. This loop contains the $$10 \, \mathrm{V}$$ source:

$$60 I_2 + 5(I_2 + I_g) = 10$$

$$65 I_2 + 5 I_g = 10 \;\Rightarrow\; 13 I_2 + I_g = 2 \quad\text{...(c)}$$

Solving. Multiply (b) by 10: $$\;20 I_1 - 60 I_g - 10 I_2 = 0$$ ...(d).
Subtract (a) from (d):

$$(20 I_1 - 60 I_g - 10 I_2) - (20 I_1 + 3 I_g - 12 I_2) = 0$$

$$-63 I_g + 2 I_2 = 0 \quad\Rightarrow\quad I_2 = 31.5 \, I_g$$

Substitute $$I_2 = 31.5 \, I_g$$ into (c):

$$13 (31.5 \, I_g) + I_g = 2$$

$$409.5 \, I_g + I_g = 2 \quad\Rightarrow\quad 410.5 \, I_g = 2$$

$$I_g = \dfrac{2}{410.5} \approx 4.87 \times 10^{-3} \, \mathrm{A}$$

The maximum current is drawn when the external resistance is zero (the terminals short-circuited), so the only resistance limiting the current is the internal resistance $$r$$ of the battery.

From $$\varepsilon = I(R + r)$$ with $$R = 0$$:

$$I_{max} = \dfrac{\varepsilon}{r} = \dfrac{12 \, \mathrm{V}}{0.4 \, \Omega} = 30 \, \mathrm{A}$$

(In practice such a large current would damage the battery; it cannot be drawn safely for long.)

For a cell of emf $$\varepsilon$$ and internal resistance $$r$$ connected to an external resistance $$R$$, the current is

$$I = \dfrac{\varepsilon}{R + r}$$

Rearranging to find $$R$$:

$$R + r = \dfrac{\varepsilon}{I} = \dfrac{10}{0.5} = 20 \, \Omega$$

$$R = 20 - r = 20 - 3 = 17 \, \Omega$$

The terminal voltage of the battery is the potential difference across the external resistor:

$$V = I R = 0.5 \times 17 = 8.5 \, \mathrm{V}$$

Equivalently, $$V = \varepsilon - I r = 10 - 0.5 \times 3 = 8.5 \, \mathrm{V}$$ — the terminal voltage is less than the emf because of the drop $$Ir$$ across the internal resistance.

The resistance varies with temperature as

$$R = R_0\left[1 + \alpha (T - T_0)\right]$$

where $$R_0 = 100 \, \Omega$$ at $$T_0 = 27.0 \, °\mathrm{C}$$, $$R = 117 \, \Omega$$ and $$\alpha = 1.70 \times 10^{-4} \, °\mathrm{C}^{-1}$$. Solving for the temperature $$T$$:

$$T - T_0 = \dfrac{R - R_0}{R_0 \, \alpha} = \dfrac{117 - 100}{100 \times 1.70 \times 10^{-4}}$$

$$T - T_0 = \dfrac{17}{1.70 \times 10^{-2}} = 1000 \, °\mathrm{C}$$

$$T = 27 + 1000 = 1027 \, °\mathrm{C}$$

The resistance of a uniform wire is related to its resistivity by

$$R = \dfrac{\rho \, l}{A} \quad\Rightarrow\quad \rho = \dfrac{R \, A}{l}$$

Substituting $$R = 5.0 \, \Omega$$, $$A = 6.0 \times 10^{-7} \, \mathrm{m^2}$$ and $$l = 15 \, \mathrm{m}$$:

$$\rho = \dfrac{5.0 \times (6.0 \times 10^{-7})}{15} = \dfrac{30 \times 10^{-7}}{15}$$

$$\rho = 2.0 \times 10^{-7} \, \Omega\,\mathrm{m}$$

The resistance at temperature $$T_2$$ is related to that at $$T_1$$ through the temperature coefficient $$\alpha$$:

$$R_2 = R_1\left[1 + \alpha (T_2 - T_1)\right]$$

so that

$$\alpha = \dfrac{R_2 - R_1}{R_1 (T_2 - T_1)}$$

Here $$R_1 = 2.1 \, \Omega$$ at $$T_1 = 27.5 \, °\mathrm{C}$$ and $$R_2 = 2.7 \, \Omega$$ at $$T_2 = 100 \, °\mathrm{C}$$. Substituting:

$$\alpha = \dfrac{2.7 - 2.1}{2.1 \times (100 - 27.5)} = \dfrac{0.6}{2.1 \times 72.5}$$

$$\alpha = \dfrac{0.6}{152.25} \approx 3.9 \times 10^{-3} \, °\mathrm{C}^{-1}$$

The resistance of the element is found from Ohm's law at the two stages.

Cold stage (room temperature $$T_1 = 27.0 \, °\mathrm{C}$$): the initial current is $$3.2 \, \mathrm{A}$$, so

$$R_1 = \dfrac{V}{I_1} = \dfrac{230}{3.2} = 71.87 \, \Omega$$

Hot stage (steady temperature $$T_2$$): the current settles to $$2.8 \, \mathrm{A}$$, so

$$R_2 = \dfrac{V}{I_2} = \dfrac{230}{2.8} = 82.14 \, \Omega$$

Using $$R_2 = R_1\left[1 + \alpha (T_2 - T_1)\right]$$:

$$T_2 - T_1 = \dfrac{R_2 - R_1}{R_1 \, \alpha} = \dfrac{82.14 - 71.87}{71.87 \times 1.70 \times 10^{-4}}$$

$$T_2 - T_1 = \dfrac{10.27}{1.22 \times 10^{-2}} \approx 840 \, °\mathrm{C}$$

$$T_2 = 27 + 840 = 867 \, °\mathrm{C}$$

The network of Fig. 3.20 is a Wheatstone-bridge type arrangement. Label the corners A (left), B (top), C (right) and D (bottom). The arms are $$AB = 10 \, \Omega$$, $$BC = 5 \, \Omega$$, $$AD = 5 \, \Omega$$, $$DC = 10 \, \Omega$$, and the bridge arm $$BD = 5 \, \Omega$$. A $$10 \, \mathrm{V}$$ battery is connected across A and C through a series resistor of $$10 \, \Omega$$.

Check for balance. A balanced bridge would require $$\dfrac{AB}{BC} = \dfrac{AD}{DC}$$. Here $$\dfrac{10}{5} = 2$$ but $$\dfrac{5}{10} = 0.5$$, so the bridge is unbalanced — a current flows through BD, and Kirchhoff's rules must be used.

Labelling the currents. Let the total current $$I$$ enter at A and split into $$x$$ through AB and $$(I - x)$$ through AD. Let $$g$$ be the current in the bridge arm BD (taken from B to D). By the junction rule the current in BC is $$x - g$$ and in DC is $$I - x + g$$.

Loop ABDA. Applying the loop rule:

$$10x + 5g - 5(I - x) = 0 \;\Rightarrow\; 15x + 5g - 5I = 0 \;\Rightarrow\; 3x + g = I \quad\text{...(1)}$$

Loop BCDB.

$$5(x - g) - 10(I - x + g) - 5g = 0$$

$$15x - 20g - 10I = 0 \;\Rightarrow\; 3x - 4g - 2I = 0 \quad\text{...(2)}$$

Substituting $$I = 3x + g$$ from (1) into (2):

$$3x - 4g - 2(3x + g) = 0 \;\Rightarrow\; -3x - 6g = 0 \;\Rightarrow\; x = -2g$$

So $$g = -\dfrac{x}{2}$$, and therefore $$I = 3x + g = 3x - \dfrac{x}{2} = \dfrac{5x}{2}$$.

Outer loop (battery → 10 Ω → A → B → C). Applying the loop rule to the $$10 \, \mathrm{V}$$ source:

$$10 = 10 I + 10x + 5(x - g)$$

$$10 = 10\left(\dfrac{5x}{2}\right) + 10x + 5\left(x + \dfrac{x}{2}\right) = 25x + 10x + 7.5x = 42.5\,x$$

$$x = \dfrac{10}{42.5} = \dfrac{4}{17} \, \mathrm{A} \approx 0.235 \, \mathrm{A}$$

Hence the total current is

$$I = \dfrac{5x}{2} = \dfrac{5}{2}\times\dfrac{4}{17} = \dfrac{10}{17} \, \mathrm{A} \approx 0.588 \, \mathrm{A}$$

and the bridge current is $$g = -\dfrac{x}{2} = -\dfrac{2}{17} \, \mathrm{A}$$; the minus sign shows it actually flows from D to B.

Current in each branch.

As a check, the voltage across the bridge network (A to C) is $$V_{AC} = 10 - 10 I = 10 - \dfrac{100}{17} = \dfrac{70}{17} \, \mathrm{V}$$, giving an equivalent resistance $$\dfrac{V_{AC}}{I} = 7 \, \Omega$$; with the series $$10 \, \Omega$$ the total is $$17 \, \Omega$$, consistent with $$I = \dfrac{10}{17} \, \mathrm{A}$$.

During charging, the $$120 \, \mathrm{V}$$ dc supply drives current into the battery, i.e. against the battery's own emf. The two emf's therefore oppose each other, so the effective emf driving the current is

$$\varepsilon_{net} = 120 - 8.0 = 112 \, \mathrm{V}$$

The total resistance in the circuit is the series resistor plus the battery's internal resistance:

$$R_{total} = 15.5 + 0.5 = 16 \, \Omega$$

So the charging current is

$$I = \dfrac{\varepsilon_{net}}{R_{total}} = \dfrac{112}{16} = 7.0 \, \mathrm{A}$$

Terminal voltage during charging. While the battery is being charged the current flows into its positive terminal, so the terminal voltage exceeds the emf by the drop $$Ir$$ across the internal resistance:

$$V = \varepsilon + I r = 8.0 + 7.0 \times 0.5 = 11.5 \, \mathrm{V}$$

(Check: the terminal voltage also equals the supply voltage minus the drop across the series resistor, $$V = 120 - I \times 15.5 = 120 - 108.5 = 11.5 \, \mathrm{V}$$.)

Purpose of the series resistor. It limits the charging current to a safe value. Without it the current would be $$\dfrac{112}{0.5} = 224 \, \mathrm{A}$$, which would severely overheat and damage the battery. The series resistor keeps the charging current small and protects the battery.

The drift speed of the conduction electrons is related to the current by $$I = n e A v_d$$, so

$$v_d = \dfrac{I}{n e A}$$

The time taken by an electron to drift the whole length $$l$$ of the wire is

$$t = \dfrac{l}{v_d} = \dfrac{l \, n \, e \, A}{I}$$

Substituting $$l = 3.0 \, \mathrm{m}$$, $$n = 8.5 \times 10^{28} \, \mathrm{m^{-3}}$$, $$e = 1.6 \times 10^{-19} \, \mathrm{C}$$, $$A = 2.0 \times 10^{-6} \, \mathrm{m^2}$$ and $$I = 3.0 \, \mathrm{A}$$:

$$t = \dfrac{(3.0)(8.5 \times 10^{28})(1.6 \times 10^{-19})(2.0 \times 10^{-6})}{3.0}$$

The factor $$3.0$$ cancels:

$$t = (8.5 \times 10^{28})(1.6 \times 10^{-19})(2.0 \times 10^{-6})$$

$$t = (8.5 \times 1.6 \times 2.0) \times 10^{28 - 19 - 6} = 27.2 \times 10^{3} \, \mathrm{s}$$

$$t \approx 2.7 \times 10^{4} \, \mathrm{s}$$

This is about $$7.5$$ hours — a striking illustration of how slow the actual drift of the electrons is, even though the current itself flows the moment the circuit is closed.