Current Electricity

Given:
- EMF of battery, $\varepsilon = 12\,\mathrm{V}$
- Internal resistance, $r = 0.4\,\Omega$

Concept: The current drawn from a battery is maximum when the external resistance $R = 0$ (short circuit condition).

Formula:
$$I_{\max} = \frac{\varepsilon}{r}$$

Calculation:
$$I_{\max} = \frac{12}{0.4} = 30\,\mathrm{A}$$

Answer: The maximum current that can be drawn from the battery is $\boxed{30\,\mathrm{A}}$.

Given:
- EMF, $\varepsilon = 10\,\mathrm{V}$
- Internal resistance, $r = 3\,\Omega$
- Current, $I = 0.5\,\mathrm{A}$

Part (i): Finding the external resistance $R$

Using the formula:
$$I = \frac{\varepsilon}{R + r}$$
$$R + r = \frac{\varepsilon}{I} = \frac{10}{0.5} = 20\,\Omega$$
$$R = 20 - r = 20 - 3 = 17\,\Omega$$

Part (ii): Terminal voltage of the battery

The terminal voltage is the voltage across the external resistor:
$$V = \varepsilon - Ir = 10 - (0.5 \times 3) = 10 - 1.5 = 8.5\,\mathrm{V}$$

Alternatively: $V = IR = 0.5 \times 17 = 8.5\,\mathrm{V}$ ✓

Answer: Resistance of the resistor $R = 17\,\Omega$; Terminal voltage $V = 8.5\,\mathrm{V}$.

Given:
- Temperature $T_1 = 27.0^{\circ}\mathrm{C}$
- Resistance at $T_1$: $R_1 = 100\,\Omega$
- Resistance at $T_2$: $R_2 = 117\,\Omega$
- Temperature coefficient, $\alpha = 1.70 \times 10^{-4}\,^{\circ}\mathrm{C}^{-1}$

Formula:
$$R_2 = R_1\,[1 + \alpha(T_2 - T_1)]$$

Calculation:
$$117 = 100\,[1 + 1.70 \times 10^{-4}(T_2 - 27)]$$
$$\frac{117}{100} = 1 + 1.70 \times 10^{-4}(T_2 - 27)$$
$$1.17 - 1 = 1.70 \times 10^{-4}(T_2 - 27)$$
$$0.17 = 1.70 \times 10^{-4}(T_2 - 27)$$
$$T_2 - 27 = \frac{0.17}{1.70 \times 10^{-4}} = 1000^{\circ}\mathrm{C}$$
$$T_2 = 1000 + 27 = 1027^{\circ}\mathrm{C}$$

Answer: The temperature of the heating element is $\boxed{1027^{\circ}\mathrm{C}}$.

Given:
- Length of wire, $l = 15\,\mathrm{m}$
- Cross-sectional area, $A = 6.0 \times 10^{-7}\,\mathrm{m}^2$
- Resistance, $R = 5.0\,\Omega$

Formula:
$$R = \frac{\rho l}{A} \implies \rho = \frac{RA}{l}$$

Calculation:
$$\rho = \frac{5.0 \times 6.0 \times 10^{-7}}{15}$$
$$\rho = \frac{30 \times 10^{-7}}{15} = 2.0 \times 10^{-7}\,\Omega\,\mathrm{m}$$

Answer: The resistivity of the material is $\boxed{2.0 \times 10^{-7}\,\Omega\,\mathrm{m}}$.

Given:
- $R_1 = 2.1\,\Omega$ at $T_1 = 27.5^{\circ}\mathrm{C}$
- $R_2 = 2.7\,\Omega$ at $T_2 = 100^{\circ}\mathrm{C}$

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

Solving for $\alpha$:
$$\alpha = \frac{R_2 - R_1}{R_1(T_2 - T_1)}$$
$$\alpha = \frac{2.7 - 2.1}{2.1 \times (100 - 27.5)}$$
$$\alpha = \frac{0.6}{2.1 \times 72.5}$$
$$\alpha = \frac{0.6}{152.25}$$
$$\alpha \approx 3.94 \times 10^{-3}\,^{\circ}\mathrm{C}^{-1}$$

Answer: The temperature coefficient of resistivity of silver is $\alpha \approx 3.94 \times 10^{-3}\,^{\circ}\mathrm{C}^{-1}$.

Given:
- Supply voltage, $V = 230\,\mathrm{V}$
- Initial current (at room temperature), $I_1 = 3.2\,\mathrm{A}$
- Steady current, $I_2 = 2.8\,\mathrm{A}$
- Room temperature, $T_1 = 27.0^{\circ}\mathrm{C}$
- $\alpha = 1.70 \times 10^{-4}\,^{\circ}\mathrm{C}^{-1}$

Step 1: Find initial and steady resistances.

Since voltage is constant:
$$R_1 = \frac{V}{I_1} = \frac{230}{3.2} = 71.875\,\Omega$$
$$R_2 = \frac{V}{I_2} = \frac{230}{2.8} = 82.14\,\Omega$$

Step 2: Use the temperature-resistance relation.
$$R_2 = R_1[1 + \alpha(T_2 - T_1)]$$
$$T_2 - T_1 = \frac{R_2 - R_1}{\alpha R_1}$$
$$T_2 - T_1 = \frac{82.14 - 71.875}{1.70 \times 10^{-4} \times 71.875}$$
$$T_2 - T_1 = \frac{10.265}{0.012219} \approx 840^{\circ}\mathrm{C}$$

Step 3: Find steady temperature.
$$T_2 = 27 + 840 = 867^{\circ}\mathrm{C}$$

Answer: The steady temperature of the heating element is approximately $\boxed{867^{\circ}\mathrm{C}}$.

Note: The figure shows a Wheatstone-bridge-type network. Based on the standard NCERT figure 3.20, the network has the following configuration: A battery of 10 V is connected between points A and B. The resistances are: $R_{AB}$ (top-left) $= 10\,\Omega$, $R_{AD}$ (bottom-left) $= 5\,\Omega$, $R_{BC}$ (top-right) $= 5\,\Omega$, $R_{DC}$ (bottom-right) $= 10\,\Omega$, and $R_{BD}$ (middle/galvanometer branch) $= 10\,\Omega$. The emf source (10 V) is between A and C.

Applying Kirchhoff's Laws:

Let:
- $I_1$ = current through $AB$ (10 Ω)
- $I_2$ = current through $AD$ (5 Ω)
- $I_g$ = current through $BD$ (10 Ω)
- Current through $BC$ = $I_1 - I_g$
- Current through $DC$ = $I_2 + I_g$

At junction B: current in = $I_1$, current out = $(I_1 - I_g) + I_g$ ✓

Loop ABDA (clockwise):
$$10I_1 + 10I_g - 5I_2 = 0$$
$$2I_1 + 2I_g - I_2 = 0 \quad \cdots (1)$$

Loop BCDB (clockwise):
$$5(I_1 - I_g) - 10(I_2 + I_g) - 10I_g = 0$$
$$5I_1 - 5I_g - 10I_2 - 10I_g - 10I_g = 0$$
$$5I_1 - 25I_g - 10I_2 = 0$$
$$I_1 - 5I_g - 2I_2 = 0 \quad \cdots (2)$$

Loop ADCEA (main outer loop):
$$5I_2 + 10(I_2 + I_g) = 10$$
$$15I_2 + 10I_g = 10$$
$$3I_2 + 2I_g = 2 \quad \cdots (3)$$

Solving equations (1) and (2):

From (1): $I_2 = 2I_1 + 2I_g$

Substitute into (2):
$$I_1 - 5I_g - 2(2I_1 + 2I_g) = 0$$
$$I_1 - 5I_g - 4I_1 - 4I_g = 0$$
$$-3I_1 - 9I_g = 0$$
$$I_1 = -3I_g \quad \cdots (4)$$

Substitute (4) into (1):
$$I_2 = 2(-3I_g) + 2I_g = -6I_g + 2I_g = -4I_g \quad \cdots (5)$$

Substitute (5) into (3):
$$3(-4I_g) + 2I_g = 2$$
$$-12I_g + 2I_g = 2$$
$$-10I_g = 2$$
$$I_g = -0.2\,\mathrm{A}$$

From (4): $I_1 = -3(-0.2) = 0.6\,\mathrm{A}$

From (5): $I_2 = -4(-0.2) = 0.8\,\mathrm{A}$

Currents in each branch:
- Current through $AB$ (10 Ω): $I_1 = 0.6\,\mathrm{A}$
- Current through $AD$ (5 Ω): $I_2 = 0.8\,\mathrm{A}$
- Current through $BD$ (10 Ω): $I_g = -0.2\,\mathrm{A}$ (i.e., $0.2\,\mathrm{A}$ in the direction opposite to assumed)
- Current through $BC$ (5 Ω): $I_1 - I_g = 0.6 - (-0.2) = 0.8\,\mathrm{A}$
- Current through $DC$ (10 Ω): $I_2 + I_g = 0.8 + (-0.2) = 0.6\,\mathrm{A}$
- Total current from battery: $I_1 + I_2 = 0.6 + 0.8 = 1.4\,\mathrm{A}$

Answer: $I_{AB} = 0.6\,\mathrm{A}$, $I_{AD} = 0.8\,\mathrm{A}$, $I_{BD} = 0.2\,\mathrm{A}$, $I_{BC} = 0.8\,\mathrm{A}$, $I_{DC} = 0.6\,\mathrm{A}$, Total current $= 1.4\,\mathrm{A}$.

Given:
- EMF of battery being charged, $\varepsilon = 8.0\,\mathrm{V}$
- Internal resistance of battery, $r = 0.5\,\Omega$
- DC supply voltage, $V = 120\,\mathrm{V}$
- Series resistance, $R = 15.5\,\Omega$

Step 1: Find the charging current.

During charging, the battery's emf opposes the supply voltage:
$$I = \frac{V - \varepsilon}{R + r} = \frac{120 - 8.0}{15.5 + 0.5} = \frac{112}{16} = 7.0\,\mathrm{A}$$

Step 2: Find the terminal voltage of the battery during charging.

During charging, current flows into the positive terminal of the battery, so:
$$V_{\text{terminal}} = \varepsilon + Ir = 8.0 + 7.0 \times 0.5 = 8.0 + 3.5 = 11.5\,\mathrm{V}$$

Purpose of series resistor:
The series resistor limits the charging current to a safe value. Without it, the current would be excessively large (since the net voltage $120 - 8 = 112\,\mathrm{V}$ would drive current through only the small internal resistance $0.5\,\Omega$, giving $I = 224\,\mathrm{A}$), which could damage the battery and the supply.

Answer: Terminal voltage during charging $= 11.5\,\mathrm{V}$. The series resistor limits the charging current to a safe value.

Given:
- Number density of free electrons, $n = 8.5 \times 10^{28}\,\mathrm{m}^{-3}$
- Length of wire, $l = 3.0\,\mathrm{m}$
- Area of cross-section, $A = 2.0 \times 10^{-6}\,\mathrm{m}^2$
- Current, $I = 3.0\,\mathrm{A}$
- Charge of electron, $e = 1.6 \times 10^{-19}\,\mathrm{C}$

Step 1: Find the drift velocity.

Using $I = neAv_d$:
$$v_d = \frac{I}{neA}$$
$$v_d = \frac{3.0}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.0 \times 10^{-6}}$$
$$v_d = \frac{3.0}{8.5 \times 1.6 \times 2.0 \times 10^{28-19-6}}$$
$$v_d = \frac{3.0}{27.2 \times 10^{3}}$$
$$v_d = \frac{3.0}{2.72 \times 10^{4}} \approx 1.103 \times 10^{-4}\,\mathrm{m\,s^{-1}}$$

Step 2: Find the time to drift the full length.

$$t = \frac{l}{v_d} = \frac{3.0}{1.103 \times 10^{-4}}$$
$$t \approx 2.7 \times 10^{4}\,\mathrm{s}$$

Answer: An electron takes approximately $\boxed{2.7 \times 10^{4}\,\mathrm{s}}$ (about 7.5 hours) to drift from one end of the wire to the other.