To determine the resistance of a carbon resistor with color bands brown, black, green, and gold, use the color code for resistors:

1. Color code values:

• Brown:
  $1$ 

  (1st digit)

• Black:
  $0$ 

  (2nd digit)

• Green:
  $10^5$ 

  (multiplier)

• Gold:
  $\pm 5\%$ 

  (tolerance)

2. Calculate resistance:

The resistance is calculated as:

$$R = (\text{1st digit}\,\times\,10 + \text{2nd digit}) \,\times\, \text{multiplier}.$$

Substitute values:

$$R = (1 \times 10 + 0) \times 10^5 = 10 \times 10^5 = 10^6 \, \Omega.$$

This is equal to:

$$R = 1 \, \text{M}\Omega \, (\text{megaohm}).$$

3. Tolerance:

The gold band indicates a tolerance of

$\pm 5\%$

, so the resistance can vary between:

$$1 \, \text{M}\Omega \pm 5\%.$$

Final Answer:

The resistance is:

$$\boxed{1 \, \text{M}\Omega \, \pm 5\%.}$$

To find the equivalent emf (

$E_{\text{eq}}$

) and internal resistance (

$r_{\text{eq}}$

) of two cells connected in parallel, we use the following principles:

1. Equivalent emf ( $E_{\text{eq}}$

):

In parallel connection, the total current is the sum of the currents through each cell. Using Kirchhoff's Voltage Law, the equivalent emf is given by:

$$E_{\text{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}.$$

2. Equivalent internal resistance ( $r_{\text{eq}}$

):

For resistances in parallel, the equivalent resistance is given by:

$$\frac{1}{r_{\text{eq}}} = \frac{1}{r_1} + \frac{1}{r_2}.$$

Simplify:

$$r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2}.$$

Final Answer:

The equivalent emf and internal resistance of the parallel combination are:

$$\boxed{E_{\text{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}, \quad r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2}}.$$

Using \( E = I(R + r) \), we set up equations: \( E = 0.5(2 + r) \) and \( E = 0.25(5 + r) \). Equating them gives \( r = 1\ \Omega \), which yields \( E = 0.5(2 + 1) = 1.5\text{ V} \).

The potential difference is the equivalent EMF: \(E_{\text{eq}} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{18/3 + 12/2}{1/3 + 1/2} = \frac{12}{5/6} = 14.4\text{ V}\).

Statement (A) is correct as EMF is indeed a potential difference under open circuit. Statement (B) is incorrect because during charging, the terminal potential difference is \(V = E + Ir\), which is greater than the EMF \(E\).

Assertion (A) is true. For two identical cells (emf \(E\), internal resistance \(r\)), in series, \(E_{\text{eq}} = 2E\) and \(r_{\text{eq}} = 2r\). Max power \(P_{\text{max,s}} = (2E)^2 / (4 \cdot 2r) = E^2 / (2r)\). In parallel, \(E_{\text{eq}} = E\) and \(r_{\text{eq}} = r/2\). Max power \(P_{\text{max,p}} = E^2 / (4 \cdot r/2) = E^2 / (2r)\). The maximum power is indeed the same. Reason (R) is false. For series, the load resistance for max power is \(R_L = 2r\), while for parallel, it is \(R_L = r/2\). These are not the same.

Assertion (A) is true: When a battery is being charged, the terminal voltage is \(V = \varepsilon + Ir\), which is greater than its emf \(\varepsilon\). Reason (R) is also true, as it correctly describes the terminal voltage when the battery is discharging (current is taken from it). However, (R) is not the correct explanation for (A), as (A) refers to a charging scenario (where \(V > \varepsilon\)), while (R) refers to a discharging scenario (where \(V < \varepsilon\)).

Assertion (A) is false: When a battery is supplying power (discharging), positive charge moves from the negative terminal to the positive terminal inside the battery, driven by non-electrostatic (chemical) forces. The internal electrostatic field points from the positive terminal to the negative terminal. Thus, the electrostatic force on positive ions opposes their motion, doing negative work.

Reason (R) is also considered false in this context: While the *electrostatic* field due to charge separation points from positive to negative, the effective field that *drives* the current (the non-conservative EMF field) is directed from the negative to the positive electrode. If 'Electric field' in (R) implies the driving force for current, then (R) is false. Therefore, both (A) and (R) are false.

Current in a resistor always flows from a region of higher potential to a region of lower potential. Therefore, if \(V_b > V_a\), current flows from \(b\) to \(a\). So, Assertion (A) is true. Inside a battery, the chemical processes drive positive charges from the negative terminal to the positive terminal. So, Reason (R) is true. However, the direction of current flow in a resistor (Assertion) is a fundamental concept of Ohm's law and potential difference, unrelated to the internal operation of a battery (Reason). Thus, R does not explain A.

It is a common observation that starting a car engine is more difficult on a cold day due to several factors including increased viscosity of engine oil and reduced battery performance. So, Assertion (A) is true. On a cold day, the chemical reactions inside a battery slow down, leading to an increase in its internal resistance and a decrease in its effective capacity, rather than an increase in EMF. The EMF generally tends to slightly decrease or remain similar. So, Reason (R) is false.