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:

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

When

$N$

identical cells, each of emf

$e$

and internal resistance

$r$

, are connected in series and

$n$

cells are connected in reverse, the resulting emf and internal resistance of the battery can be determined as follows:

1. Resultant emf ( $\varepsilon_0$

):

• For correctly connected cells, the total emf is:
  $$\varepsilon_{\text{correct}} = (N - n)e.$$ 
• For n reversed cells, their emf opposes the total emf. The opposing emf is:
  $$\varepsilon_{\text{reverse}} = ne.$$ 

  The net emf of the resulting battery is: 

  $$\varepsilon_0 = \varepsilon_{\text{correct}} - \varepsilon_{\text{reverse}} = (N - n)e - ne = (N - 2n)e.$$ 

2. Resultant internal resistance ( $r_0$

):

• All cells, whether correctly or incorrectly connected, contribute to the total internal resistance because resistances add in series. The total internal resistance is:
  $$r_0 = N r.$$ 

Final Answer:

The emf and internal resistance of the resulting battery are:

$$\boxed{\varepsilon_0 = (N - 2n)e, \quad r_0 = Nr.}$$

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:

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

We are tasked to find the external resistance

$R$

when the potential difference across the first cell is zero. Let the emf of each cell be

$E$

, the internal resistances of the two cells be

$r_1$

and

$r_2$

, and the external resistance be

$R$

.

Key points:

1. Current in the circuit: The total resistance in the circuit is
   $r_1 + r_2 + R$ 

   . The current $I$ 

   is given by: 

   $$I = \frac{E + E}{r_1 + r_2 + R} = \frac{2E}{r_1 + r_2 + R}.$$ 

2. Potential difference across the first cell: The potential difference across the first cell is: 

   $$V_1 = E - I r_1.$$Since the potential difference across the first cell is zero, we set

   $V_1 = 0$:

    

   $$0 = E - I r_1.$$Substitute

   $I = \frac{2E}{r_1 + r_2 + R}$:

    

   $$0 = E - \frac{2E}{r_1 + r_2 + R} \cdot r_1.$$ 

3. Simplify the equation: Rearrange: 

   $$E = \frac{2E r_1}{r_1 + r_2 + R}.$$Divide through by

   $E$(since

   $E \neq 0$):

    

   $$1 = \frac{2r_1}{r_1 + r_2 + R}.$$Multiply both sides by

   $r_1 + r_2 + R$:

    

   $$r_1 + r_2 + R = 2r_1.$$Simplify:

    

   $$R = 2r_1 - r_1 - r_2.$$ 

   $$R = r_1 - r_2.$$ 

Final Answer:

The value of

$R$

is:

$$\boxed{R = r_1 - r_2}.$$

Here 4 ohm resistor is short circuited. So current in circuit is 10/1 = 10 Ampere.

Potential V= E -ir= 10 - 10×1 = 0 (zero)

To find the internal resistance (

$r$

) of the battery, we use the following equations based on the given information:

1. Case 1: Current flows from negative to positive terminal

The potential difference is:

$$V_1 = E - I_1 r$$

Substitute

$V_1 = 10 \, \text{V}$

,

$I_1 = 3 \, \text{A}$

:

$$10 = E - 3r \tag{1}$$

2. Case 2: Current flows in reverse (from positive to negative terminal)

The potential difference is:

$$V_2 = E + I_2 r$$

Substitute

$V_2 = 15 \, \text{V}$

,

$I_2 = 2 \, \text{A}$

:

$$15 = E + 2r \tag{2}$$

3. Solve the two equations

From Equation (1):

$$E = 10 + 3r$$

Substitute

$E = 10 + 3r$

into Equation (2):

$$15 = (10 + 3r) + 2r$$

Simplify:

$$15 = 10 + 5r$$

$$5r = 5 \implies r = 1 \, \Omega$$

Final Answer:

The internal resistance of the battery is:

$$\boxed{1 \, \Omega}$$

In the given circuit of

$N$

cells, the emf (

$E_N$

) and internal resistance (

$r_N$

) are related as

$E_N = 1.5r_N$

. To find the current (

$I$

), we follow these steps:

1. Equivalent emf and resistance:
   • The cells are connected in series, so:
     $$E_{\text{eq}} = E_N = 1.5r_N, \quad r_{\text{eq}} = r_N.$$ 
2. Total resistance:
   • Let
     $R_{\text{ext}}$ 

     be the external resistance of the circuit. From the figure, the circuit resistance is: $$R_{\text{total}} = r_N + R_{\text{ext}}.$$ 

3. Ohm's Law:
   • The current in the circuit is:
     $$I = \frac{E_{\text{eq}}}{R_{\text{total}}} = \frac{1.5r_N}{r_N + R_{\text{ext}}}.$$ 
4. Condition for ( I = 1.5 , \text{A}:
   • Substituting
     $I = 1.5$ 

     into the equation: 

     $$1.5 = \frac{1.5r_N}{r_N + R_{\text{ext}}}.$$ 

   • Simplifying: 

     $$r_N + R_{\text{ext}} = r_N \implies R_{\text{ext}} = 0.$$ 

Thus, the current

$I = 1.5 \, \text{A}$

when

$R_{\text{ext}} = 0$

, meaning there is no external resistance.

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 \( E = 0.5(2 + 1) = 1.5\text{ V} \).

Statements (A), (B), and (D) are conceptually correct definitions and behaviors of EMF and cells. Statement (C) is incorrect because series cells increase total EMF.

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