Current Electricity — NCERT Solutions

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.