Current ElectricityClass 12 Physics NCERT Solutions

The maximum current that can be drawn from a battery occurs when the external resistance in the circuit is zero. This situation effectively short-circuits the battery.

The formula for the current (I) drawn from a battery is given by: I = ε / (R + r)

where R is the external resistance.

To find the maximum current (I_max), we set the external resistance R = 0: I_max = ε / (0 + r) = ε / r

Substituting the given values: I_max = 12 V / 0.4 Ω I_max = 30 A

Thus, the maximum current that can be drawn from the battery is 30 A.

Here we need to find the external resistance (R) and the terminal voltage (V) of the battery.

1. To find the resistance of the resistor (R): The relationship between emf, current, and total resistance is given by Ohm's law for the entire circuit: I = ε / (R + r)

Rearranging the formula to solve for R: R + r = ε / I R = (ε / I) - r

Substituting the given values: R = (10 V / 0.5 A) - 3 Ω R = 20 Ω - 3 Ω R = 17 Ω

So, the resistance of the resistor is 17 Ω.

2. To find the terminal voltage of the battery (V): The terminal voltage is the potential difference across the external resistor, or the potential difference across the terminals of the battery when current is flowing. It can be calculated in two ways:

Both methods yield the same result. The terminal voltage of the battery when the circuit is closed is 8.5 V.

The relationship between resistance and temperature for a conductor is given by the formula: R_T = R_0 [1 + α(T - T_0)]

We need to find the final temperature, T.

Let us rearrange the formula to solve for T: R_T / R_0 = 1 + α(T - T_0) (R_T / R_0) - 1 = α(T - T_0) (R_T - R_0) / R_0 = α(T - T_0) T - T_0 = (R_T - R_0) / (R_0 * α) T = T_0 + [(R_T - R_0) / (R_0 * α)]

Now, substitute the given values into the equation: T = 27.0 + [(117 - 100) / (100 * 1.70 × 10⁻⁴)] T = 27.0 + [17 / (1.70 × 10⁻²)] T = 27.0 + [17 / 0.017] T = 27.0 + 1000 T = 1027 °C

Therefore, the temperature of the element is 1027 °C when its resistance is 117 Ω.

The resistance (R) of a wire is related to its length (l), cross-sectional area (A), and the resistivity (ρ) of its material by the formula: R = ρ * (l / A)

We need to find the resistivity (ρ) of the material.

Rearranging the formula to solve for ρ: ρ = R * (A / l)

Now, substitute the given values into the equation: ρ = 5.0 Ω * (6.0 × 10⁻⁷ m² / 15 m) ρ = (30.0 × 10⁻⁷) / 15 Ω·m ρ = 2.0 × 10⁻⁷ Ω·m

Thus, the resistivity of the material at the temperature of the experiment is 2.0 × 10⁻⁷ Ω·m.

The relationship between resistance and temperature is given by: R_2 = R_1 [1 + α(T_2 - T_1)] where α is the temperature coefficient of resistivity (or resistance).

We need to determine the temperature coefficient of resistivity, α.

Rearranging the formula to solve for α: R_2 / R_1 = 1 + α(T_2 - T_1) (R_2 / R_1) - 1 = α(T_2 - T_1) (R_2 - R_1) / R_1 = α(T_2 - T_1) α = (R_2 - R_1) / [R_1 * (T_2 - T_1)]

Now, substitute the given values into the equation: α = (2.7 Ω - 2.1 Ω) / [2.1 Ω * (100 °C - 27.5 °C)] α = 0.6 / [2.1 * (72.5)] α = 0.6 / 152.25 α ≈ 0.00394 °C⁻¹

Therefore, the temperature coefficient of resistivity of silver is approximately 0.00394 °C⁻¹.

This problem involves calculating the final temperature based on the change in resistance, which is inferred from the change in current.

Step 1: Calculate the initial and final resistances. The initial current is drawn when the element is at room temperature. The initial resistance (R_1) can be calculated using Ohm's law: R_1 = V / I_1 = 230 V / 3.2 A = 71.875 Ω

As the element heats up, its resistance increases, causing the current to decrease to a steady value. The steady (final) resistance (R_2) at the high temperature (T_2) is: R_2 = V / I_2 = 230 V / 2.8 A ≈ 82.143 Ω

Step 2: Use the temperature-resistance formula to find the final temperature (T_2). The formula relating resistance and temperature is: R_2 = R_1 [1 + α(T_2 - T_1)]

Rearranging to solve for T_2: T_2 - T_1 = (R_2 - R_1) / (R_1 * α) T_2 = T_1 + [(R_2 - R_1) / (R_1 * α)]

Step 3: Substitute the calculated and given values. T_2 = 27.0 + [(82.143 - 71.875) / (71.875 * 1.70 × 10⁻⁴)] T_2 = 27.0 + [10.268 / (0.01221875)] T_2 = 27.0 + 840.35 T_2 ≈ 867.4 °C

Thus, the steady temperature of the heating element is approximately 867.4 °C.

To solve this network, we will apply Kirchhoff's rules. Let the current from the 10V cell be I. At junction A, this current splits.

At junction B, current I_1 splits.

At junction D, currents (I - I_1) and I_g combine.

At junction C, currents from BC and DC combine to give the total current I, which returns to the cell.

Now, we apply Kirchhoff's loop rule to three independent closed loops:

1. Loop ABDA: -10(I_1) - 5(I_g) + 5(I - I_1) = 0 -10I_1 - 5I_g + 5I - 5I_1 = 0 -15I_1 - 5I_g + 5I = 0 Dividing by 5: -3I_1 - I_g + I = 0 => I = 3I_1 + I_g (Equation 1)

2. Loop BCDB: -5(I_1 - I_g) + 10(I - I_1 + I_g) + 5(I_g) = 0 -5I_1 + 5I_g + 10I - 10I_1 + 10I_g + 5I_g = 0 -15I_1 + 20I_g + 10I = 0 Dividing by 5: -3I_1 + 4I_g + 2I = 0 (Equation 2)

3. Loop ADCE (outer loop including the cell): -5(I - I_1) - 10(I - I_1 + I_g) + 10 = 0 -5I + 5I_1 - 10I + 10I_1 - 10I_g + 10 = 0 -15I + 15I_1 - 10I_g + 10 = 0 Dividing by 5: -3I + 3I_1 - 2I_g + 2 = 0 (Equation 3)

Now we solve these three simultaneous equations. Substitute Equation 1 into Equation 2: -3I_1 + 4I_g + 2(3I_1 + I_g) = 0 -3I_1 + 4I_g + 6I_1 + 2I_g = 0 3I_1 + 6I_g = 0 => I_1 = -2I_g

Substitute I and I_1 in terms of I_g into Equation 3: I = 3(-2I_g) + I_g = -6I_g + I_g = -5I_g -3(-5I_g) + 3(-2I_g) - 2I_g + 2 = 0 15I_g - 6I_g - 2I_g + 2 = 0 7I_g + 2 = 0 I_g = -2/7 A

The negative sign indicates that the current in branch BD flows from D to B.

Now, we can find the other currents:

Finally, we find the current in each branch:

When a battery is being charged, the external DC supply must provide a voltage greater than the battery's emf. The supply pushes current into the positive terminal of the battery, against its emf.

The net effective voltage in the circuit that drives the current is the difference between the supply voltage and the battery's emf, because the emf opposes the charging current. Effective Voltage, V_eff = V_supply - ε = 120 V - 8.0 V = 112 V

The total resistance in the circuit is the sum of the series resistor and the internal resistance of the battery. Total Resistance, R_total = R + r = 15.5 Ω + 0.5 Ω = 16.0 Ω

The charging current (I) in the circuit is: I = V_eff / R_total = 112 V / 16.0 Ω = 7 A

During charging, the terminal voltage (V_terminal) across the battery is the sum of its emf and the voltage drop across its internal resistance. V_terminal = ε + I * r V_terminal = 8.0 V + (7 A * 0.5 Ω) V_terminal = 8.0 V + 3.5 V V_terminal = 11.5 V

So, the terminal voltage of the battery during charging is 11.5 V.

The series resistor is crucial for controlling the charging current. If the series resistor were not present (R=0), the total resistance in the circuit would only be the battery's internal resistance (r = 0.5 Ω). The charging current would then be: I = (120 V - 8.0 V) / 0.5 Ω = 112 V / 0.5 Ω = 224 A

This extremely large current would cause excessive heating (due to power loss P = I²r) and would likely damage the battery permanently. The purpose of the series resistor is to limit the charging current to a safe and manageable value (7 A in this case), preventing overheating and damage to the battery.

To find the time taken for an electron to drift the length of the wire, we first need to calculate the drift velocity (v_d) of the electrons.

Step 1: Calculate the drift velocity (v_d). The relationship between current (I) and drift velocity (v_d) is given by the formula: I = n * A * e * v_d

Rearranging the formula to solve for v_d: v_d = I / (n * A * e)

Substituting the given values: v_d = 3.0 / (8.5 × 10²⁸ * 2.0 × 10⁻⁶ * 1.6 × 10⁻¹⁹) v_d = 3.0 / (8.5 * 2.0 * 1.6 * 10^(28 - 6 - 19)) v_d = 3.0 / (27.2 * 10³) v_d ≈ 0.1103 × 10⁻³ m/s = 1.103 × 10⁻⁴ m/s

Step 2: Calculate the time (t) taken to drift. The time taken for an electron to drift a distance L at a constant drift velocity v_d is: t = L / v_d

Substituting the values: t = 3.0 m / (1.103 × 10⁻⁴ m/s) t ≈ 2.72 × 10⁴ s

To make this value more understandable, we can convert it to hours: t (in hours) = (2.72 × 10⁴ s) / (3600 s/hour) t ≈ 7.56 hours

Therefore, it takes an electron approximately 2.72 × 10⁴ seconds (or about 7.6 hours) to drift from one end of the 3.0 m long wire to the other.