Chapter 3: Current Electricity (Current Electricity)

The current is largest when the external resistance is zero (terminals shorted), so only the internal resistance limits it:

$$I_{\max }=\frac{\epsilon }{r}=\frac{12}{0.4}=30\text{A}$$

30 A.

$I=\frac{\epsilon }{R+r}$, so $R+r=\frac{10}{0.5}=20$ Ω and $R=20-3=17$ Ω.

Terminal voltage: $V=IR=0.5\times 17=8.5$ V (check: $\epsilon -Ir=10-0.5\times 3=8.5$ V).

R = 17 Ω; terminal voltage 8.5 V.

$R_T=R_0[1+\alpha (T-T_0)]$:

$$117=100[1+1.70\times 10^{-4}(T-27)]\Rightarrow T-27=\frac{0.17}{1.70\times 10^{-4}}=1000{}^{\circ }\text{C}$$

So $T=1027$ °C.

About 1027 °C.

$$\rho =\frac{RA}{l}=\frac{5.0\times 6.0\times 10^{-7}}{15}=2.0\times 10^{-7}\Omega \text{m}$$

(The current is kept small so that the wire does not heat up and change its resistance.)

$\rho =2.0\times 10^{-7}$ Ω m.

$$\alpha =\frac{R_2-R_1}{R_1(T_2-T_1)}=\frac{2.7-2.1}{2.1\times (100-27.5)}=\frac{0.6}{152.25}\approx 3.94\times 10^{-3}{}^{\circ }{\text{C}}^{-1}$$

$\alpha \approx 3.94\times 10^{-3}$ °C⁻¹.

At room temperature: $R_1=\frac{230}{3.2}=71.9$ Ω. When hot and steady: $R_2=\frac{230}{2.8}=82.1$ Ω.

$$T_2-T_1=\frac{R_2-R_1}{R_1\alpha }=\frac{82.1-71.9}{71.9\times 1.70\times 10^{-4}}\approx 840{}^{\circ }\text{C}$$

So $T_2\approx 840+27=867$ °C.

About 867 °C.

The network: AB = 10 Ω, BC = 5 Ω, AD = 5 Ω, DC = 10 Ω, BD = 5 Ω, and a 10 V cell with a 10 Ω resistor connected between A and C.

Equivalent resistance between A and C. Take $V_C=0$ and $V_A=V$. Kirchhoff's junction rule at B and D:

At B: $\frac{V-V_B}{10}+\frac{V_D-V_B}{5}-\frac{V_B}{5}=0$; at D: $\frac{V-V_D}{5}+\frac{V_B-V_D}{5}-\frac{V_D}{10}=0$.

Solving: $V_B=\frac{3}{7}V$ and $V_D=\frac{4}{7}V$. The current leaving A is $\frac{V-V_B}{10}+\frac{V-V_D}{5}=\frac{V}{7}$, so $R_{AC}=7$ Ω.

Total current. $I=\frac{10}{10+7}=\frac{10}{17}$ A, and $V=V_A=7\times \frac{10}{17}=\frac{70}{17}$ V.

Branch currents:

• AB: $\frac{V-V_B}{10}=\frac{4}{7}\cdot \frac{70}{17}\cdot \frac{1}{10}=\frac{4}{17}$ A
• AD: $\frac{V-V_D}{5}=\frac{3}{7}\cdot \frac{70}{17}\cdot \frac{1}{5}=\frac{6}{17}$ A
• DB: $\frac{V_D-V_B}{5}=\frac{1}{7}\cdot \frac{70}{17}\cdot \frac{1}{5}=\frac{2}{17}$ A, from D to B
• BC: $\frac{V_B}{5}=\frac{6}{17}$ A
• DC: $\frac{V_D}{10}=\frac{4}{17}$ A
• Through the cell and the 10 Ω resistor: $\frac{10}{17}$ A

Check at C: $\frac{6}{17}+\frac{4}{17}=\frac{10}{17}$ A. ✔

AB 4/17 A, AD 6/17 A, DB 2/17 A (D to B), BC 6/17 A, DC 4/17 A, main branch 10/17 A.

While charging, the supply drives current against the battery's emf:

$$I=\frac{120-8}{15.5+0.5}=\frac{112}{16}=7\text{A}$$

Terminal voltage while charging: $V=\epsilon +Ir=8+7\times 0.5=11.5$ V.

Purpose of the series resistor: it limits the charging current. Without it the current would be $\frac{112}{0.5}=224$ A, which would overheat and damage the battery.

11.5 V; the series resistor limits the charging current to a safe value.

$$v_d=\frac{I}{neA}=\frac{3.0}{8.5\times 10^{28}\times 1.6\times 10^{-19}\times 2.0\times 10^{-6}}\approx 1.1\times 10^{-4}\text{m/s}$$

$$t=\frac{l}{v_d}=\frac{3.0}{1.1\times 10^{-4}}\approx 2.7\times 10^4\text{s}$$

That is about 7.5 hours. (The current still starts almost instantly, because the electric field travels along the wire at nearly the speed of light and sets all the electrons drifting at once.)

About $2.7\times 10^4$ s (about 7.5 h).