CBSE Class 12 Physics: Current Electricity NCERT Solutions (Chapter 3)

We are given the following information:

• Surface charge density of the earth, $\sigma = 10^{-9} \text{ C m}^{-2}$
• Current over the entire globe, $I = 1800 \text{ A}$
• Radius of the earth, $r = 6.37 \times 10^6 \text{ m}$

First, we calculate the total surface area of the Earth, which is a sphere:

$$A\; =\; 4\backslash pi\; r^2$$

Substituting the given radius:

$$A\; =\; 4\backslash pi\; \backslash times\; (6.37\; \backslash times\; 10^6\; \backslash text\{\; m\})^2$$

$$A\; \backslash approx\; 5.09\; \backslash times\; 10^\{14\}\; \backslash text\{\; m\}^2$$

Next, we find the total charge ($q$) on the Earth's surface using the surface charge density:

$$q\; =\; \backslash sigma\; \backslash times\; A$$

$$q\; =\; (10^\{-9\}\; \backslash text\{\; C\; m\}^\{-2\})\; \backslash times\; (5.09\; \backslash times\; 10^\{14\}\; \backslash text\{\; m\}^2)$$

$$q\; =\; 5.09\; \backslash times\; 10^\{5\}\; \backslash text\{\; C\}$$

The time ($t$) required to neutralize this charge is given by the relationship between current, charge, and time ($I = q/t$). We can rearrange this to solve for time:

$$t\; =\; \backslash frac\{q\}\{I\}$$

Now, we substitute the values of total charge and global current:

$$t\; =\; \backslash frac\{5.09\; \backslash times\; 10^\{5\}\; \backslash text\{\; C\}\}\{1800\; \backslash text\{\; A\}\}$$

$$t\; \backslash approx\; 282.77\; \backslash text\{\; seconds\}$$

Therefore, the time required to neutralize the Earth's surface, under these hypothetical conditions, is approximately 282.77 seconds.

We are given:

• Number of secondary cells, $n = 6$
• Emf of each cell, $E = 2.0 \text{ V}$
• Internal resistance of each cell, $r = 0.015 \Omega$
• External resistance, $R = 8.5 \Omega$

When cells are connected in series, the total emf is the sum of individual emfs, and the total internal resistance is the sum of individual internal resistances.

Total emf of the combination:

E_{total} = n \times E = 6 \times 2.0 \text{ V} = 12.0 \text{ V}

Total internal resistance of the combination:

r_{total} = n \times r = 6 \times 0.015 \Omega = 0.09 \Omega

The total resistance in the circuit is the sum of the external resistance and the total internal resistance:

R_{circuit} = R + r_{total} = 8.5 \Omega + 0.09 \Omega = 8.59 \Omega

The current drawn from the supply ($I$) is given by Ohm's law applied to the entire circuit:

I = \frac{E_{total}}{R_{circuit}} = \frac{12.0 \text{ V}}{8.59 \Omega}

I \approx 1.39 \text{ A}

The terminal voltage ($V$) of the battery is the voltage across the external resistance $R$. It can be calculated as:

V = I \times R = 1.39 \text{ A} \times 8.5 \Omega

V \approx 11.87 \text{ V}

Alternatively, the terminal voltage can be calculated as the total emf minus the voltage drop across the internal resistance: $V = E_{total} - I \times r_{total} = 12.0 \text{ V} - (1.39 \text{ A} \times 0.09 \Omega) \approx 12.0 \text{ V} - 0.125 \text{ V} \approx 11.875 \text{ V}$.

Therefore, the current drawn from the supply is approximately 1.39 A, and the terminal voltage is approximately 11.87 V.

We are given the characteristics of the aged secondary cell:

• Emf of the cell, $E = 1.9 \text{ V}$
• Internal resistance of the cell, $r = 380 \Omega$

The maximum current that can be drawn from a cell is when the external resistance is negligible (ideally zero, representing a short circuit). In this case, the current is limited only by the cell's internal resistance. Using Ohm's law:

$$I\_\{max\}\; =\; \backslash frac\{E\}\{r\}$$

Substituting the given values:

$$I\_\{max\}\; =\; \backslash frac\{1.9\; \backslash text\{\; V\}\}\{380\; \backslash Omega\}$$

$$I\_\{max\}\; =\; 0.005\; \backslash text\{\; A\}$$

The maximum current that can be drawn from this aged cell is 0.005 A.

Regarding driving the starting motor of a car: A car's starting motor requires a very large surge of current, typically hundreds of amperes, to crank the engine. Since the maximum current this cell can provide is only 0.005 A (which is 5 milliamperes), it is far too low to power a car's starting motor.

Therefore, this cell could not drive the starting motor of a car.