CBSE Class 12 Physics: Current Electricity NCERT Solutions

The electromotive force (emf) of the battery is given as $E\; =\; 12\; \backslash text\{\; V\}$. The internal resistance of the battery is given as $r\; =\; 0.4\; \backslash Omega$. The maximum current that can be drawn from a battery occurs when the external circuit resistance is zero (i.e., a short circuit). In this case, 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 negligible, the maximum current ($I\_\{max\}$) can be calculated as:

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

Substituting the given values:

$$I\_\{max\}\; =\; \backslash frac\{12\; \backslash text\{\; V\}\}\{0.4\; \backslash Omega\}\; =\; 30\; \backslash text\{\; A\}$$

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

Given:

Emf of the battery, E = 10 \text{ V}

Internal resistance of the battery, r = 3 \Omega

Current 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, the current (I) is given by:

I = \frac{E}{R + r}

We can rearrange this formula to find the total resistance (R + r):

R + r = \frac{E}{I}

Substituting the given values:

R + r = \frac{10 \text{ V}}{0.5 \text{ A}} = 20 \Omega

Now, we can find the resistance of the external resistor (R):

R = (R + r) - r = 20 \Omega - 3 \Omega = 17 \Omega

The terminal voltage (V) of the battery when the circuit is closed is the potential difference across the external resistor. It can be calculated using Ohm's law for the external resistor:

V = IR

Substituting the values:

V = (0.5 \text{ A}) \times (17 \Omega) = 8.5 \text{ V}

Alternatively, the terminal voltage can be calculated as V = E - Ir:

V = 10 \text{ V} - (0.5 \text{ A} \times 3 \Omega) = 10 \text{ V} - 1.5 \text{ V} = 8.5 \text{ V}

Therefore, the resistance of the resistor is 17 \Omega, and the terminal voltage of the battery is 8.5 V.

1. When resistors are connected in series, the total resistance of the combination 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 ($R\_\{total\}$) is:

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

   $$R\_\{total\}\; =\; 1\; \backslash Omega\; +\; 2\; \backslash Omega\; +\; 3\; \backslash Omega\; =\; 6\; \backslash Omega$$

   So, the total resistance of the combination is 6 $\backslash Omega$.

2. The combination of resistors is connected to a battery with emf E = 12 \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_{total} = 6 \Omega. The current (I) flowing through the circuit can be calculated using Ohm's law:

   I = \frac{E}{R_{total}}

   I = \frac{12 \text{ V}}{6 \Omega} = 2 \text{ A}

   Since the resistors are connected in series, the same current flows through each resistor. The potential drop across each resistor can be calculated using Ohm's law (V = IR):

   Potential drop across the first resistor (R_1 = 1 \Omega):

   V_1 = I \times R_1 = 2 \text{ A} \times 1 \Omega = 2 \text{ V}

   Potential drop across the second resistor (R_2 = 2 \Omega):

   V_2 = I \times R_2 = 2 \text{ A} \times 2 \Omega = 4 \text{ V}

   Potential drop across the third resistor (R_3 = 3 \Omega):

   V_3 = I \times R_3 = 2 \text{ A} \times 3 \Omega = 6 \text{ V}

   We can check that the sum of the potential drops equals the battery's emf: V_1 + V_2 + V_3 = 2 \text{ V} + 4 \text{ V} + 6 \text{ V} = 12 \text{ V}, which matches the emf of the battery.