Solution:

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:

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

Discuss this question

Solution:

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:

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

Discuss this question

Solution:

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:

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

Discuss this question

Solution:

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:

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

Discuss this question

Solution:

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)

Discuss this question

Solution:

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:

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

Discuss this question

Solution:

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.

Discuss this question