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