CBSE Class 12 Physics Chapter 6: Current Electricity NCERT Solutions

The electromotive force (emf) of the car's battery is given as $E\; =\; 12\; \backslash text\{\; V\}$. The internal resistance of the battery is $r\; =\; 0.4\; \backslash Omega$. The maximum current that can be drawn from a battery occurs when the external resistance connected to it is zero (i.e., a short circuit). In this scenario, the entire emf is used to drive the current through the internal resistance.

Using Ohm's law for the entire circuit, where the external resistance is assumed to be negligible:

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

Substituting the given values:

$I\_\{\backslash text\{max\}\}\; =\; \backslash frac\{12\; \backslash text\{\; V\}\}\{0.4\; \backslash Omega\}$

$I\_\{\backslash text\{max\}\}\; =\; 30\; \backslash text\{\; A\}$

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

We are given the following values for the battery and the circuit:

• Emf of the battery, $E\; =\; 10\; \backslash text\{\; V\}$
• Internal resistance of the battery, $r\; =\; 3\; \backslash Omega$
• Current flowing in the circuit, $I\; =\; 0.5\; A$

Let the resistance of the external resistor be $R$. According to Ohm's law applied to the entire circuit (including the internal resistance), the current is given by:

$$I\; =\; \backslash frac\{E\}\{R\; +\; r\}$$

To find the resistance $R$, we can rearrange the formula:

$R\; +\; r\; =\; \backslash frac\{E\}\{I\}$

$R\; +\; 3\; \backslash Omega\; =\; \backslash frac\{10\; \backslash text\{\; V\}\}\{0.5\; A\}$

$R\; +\; 3\; \backslash Omega\; =\; 20\; \backslash Omega$

$R\; =\; 20\; \backslash Omega\; -\; 3\; \backslash Omega$

$R\; =\; 17\; \backslash Omega$

So, the resistance of the resistor is 17 Ω.

Next, we need to find the terminal voltage ($V$) of the battery when the circuit is closed. The terminal voltage is the potential difference across the external resistor, which can be calculated using Ohm's law for the external resistor:

$V\; =\; IR$

Substituting the values of current and the calculated resistance:

$V\; =\; 0.5\; A\; \backslash times\; 17\; \backslash Omega$

$V\; =\; 8.5\; V$

Alternatively, the terminal voltage can also be calculated as the emf minus the voltage drop across the internal resistance:

$V\; =\; E\; -\; Ir$

$V\; =\; 10\; \backslash text\{\; V\}\; -\; (0.5\; A\; \backslash times\; 3\; \backslash Omega)$

$V\; =\; 10\; \backslash text\{\; V\}\; -\; 1.5\; \backslash text\{\; V\}$

$V\; =\; 8.5\; \backslash text\{\; V\}$

Therefore, the resistance of the resistor is 17 Ω, and the terminal voltage of the battery when the circuit is closed is 8.5 V.

(a) Total resistance in series combination:

When resistors are connected in series, the total resistance ($R\_\{\backslash text\{total\}\}$) is the sum of the individual resistances. Given the resistances are $R\_1\; =\; 1\; \backslash Omega$, $R\_2\; =\; 2\; \backslash Omega$, and $R\_3\; =\; 3\; \backslash Omega$.

The total resistance is calculated as:

$R\_\{\backslash text\{total\}\}\; =\; R\_1\; +\; R\_2\; +\; R\_3$

$R\_\{\backslash text\{total\}\}\; =\; 1\; \backslash Omega\; +\; 2\; \backslash Omega\; +\; 3\; \backslash Omega$

$R\_\{\backslash text\{total\}\}\; =\; 6\; \backslash Omega$

Thus, the total resistance of the series combination is 6 Ω.

(b) Potential drop across each resistor:

The combination of resistors is connected to a battery with an emf $E\; =\; 12\; \backslash text\{\; V\}$ and negligible internal resistance. Since the internal resistance is negligible, the total resistance of the circuit is equal to the total resistance of the combination, which is $R\_\{\backslash text\{total\}\}\; =\; 6\; \backslash Omega$.

First, we calculate the current ($I$) flowing through the circuit using Ohm's law:

$$I\; =\; \backslash frac\{E\}\{R\_\{\backslash text\{total\}\}\}$$

$I\; =\; \backslash frac\{12\; \backslash text\{\; V\}\}\{6\; \backslash Omega\}$

$I\; =\; 2\; \backslash text\{\; A\}$

Since the resistors are connected in series, the same current ($I\; =\; 2\; A$) flows through each resistor.

Now, we can find the potential drop across each resistor using Ohm's law ($V\; =\; IR$):

Potential drop across the first resistor ($R\_1\; =\; 1\; \backslash Omega$):

$V\_1\; =\; I\; R\_1\; =\; 2\; A\; \backslash times\; 1\; \backslash Omega\; =\; 2\; \backslash text\{\; V\}$

Potential drop across the second resistor ($R\_2\; =\; 2\; \backslash Omega$):

$V\_2\; =\; I\; R\_2\; =\; 2\; A\; \backslash times\; 2\; \backslash Omega\; =\; 4\; \backslash text\{\; V\}$

Potential drop across the third resistor ($R\_3\; =\; 3\; \backslash Omega$):

$V\_3\; =\; I\; R\_3\; =\; 2\; A\; \backslash times\; 3\; \backslash Omega\; =\; 6\; \backslash text\{\; V\}$

We can check that the sum of the potential drops equals the battery's emf: $V\_1\; +\; V\_2\; +\; V\_3\; =\; 2\; \backslash text\{\; V\}\; +\; 4\; \backslash text\{\; V\}\; +\; 6\; \backslash text\{\; V\}\; =\; 12\; \backslash text\{\; V\}$, which matches the given emf.