Case 1: Using the galvanometer as a voltmeter

Given Data:

• Coil resistance of galvanometer:
  $R_g = 100 \, \Omega$ 

  ,

• Full-scale deflection current of the galvanometer:
  $I_g = 50 \, \mu A = 50 \times 10^{-6} \, \text{A}$ 

  ,

• Range of the voltmeter to be designed:
  $V = 50 \, \text{V}$ 

  .

Total Resistance Required ( $R_v$

):

The total resistance

$R_v$

of the voltmeter is determined using Ohm's law:

$$R_v = \frac{V}{I_g} = \frac{50}{50 \times 10^{-6}} = 10^6 \, \Omega.$$

Since the galvanometer already has a resistance

$R_g = 100 \, \Omega$

, the additional series resistance

$R_s$

required is:

$$R_s = R_v - R_g = 10^6 - 100 \approx 10^6 \, \Omega.$$

Conclusion:

• To use the galvanometer as a voltmeter of range 50 V, the series resistance required is
  $\mathbf{10^6 \, \Omega}$ 

  (Option A is correct).

Case 2: Using the galvanometer as an ammeter

Given Data:

• Full-scale deflection current of the galvanometer:
  $I_g = 50 \, \mu A = 50 \times 10^{-6} \, \text{A}$ 

  ,

• Resistance of the galvanometer:
  $R_g = 100 \, \Omega$ 

  ,

• Range of the ammeter to be designed:
  $I = 10 \, \text{mA} = 10 \times 10^{-3} \, \text{A}$ 

  .

Shunt Resistance ( $R_s$

):

The shunt resistance is connected in parallel with the galvanometer to allow the additional current (

$I - I_g$

) to pass through it. The voltage across the galvanometer and the shunt must be equal:

$$V_g = V_s \quad \Rightarrow \quad I_g R_g = I_s R_s,$$

where

$I_s = I - I_g = 10 \times 10^{-3} - 50 \times 10^{-6} = 9.95 \times 10^{-3} \, \text{A}$

.

Using the above relation, the shunt resistance is:

$$R_s = \frac{I_g R_g}{I_s} = \frac{(50 \times 10^{-6}) \cdot 100}{9.95 \times 10^{-3}} = 0.5 \, \Omega.$$

Conclusion:

• To use the galvanometer as an ammeter of range 10 mA, the shunt resistance required is
  $\mathbf{0.5 \, \Omega}$ 

  (Option C is correct).

Final Answer:

The correct options are:

$$\boxed{\text{A and C}}$$

The current I is calculated by:

Thus, the correct answer is 0.8 A (Option 4)

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.

To solve this, we need to determine the shunt resistance (

$R_S$

) required to extend the range of the milliammeter from 10 mA to 1 A.

Key Points:

1. Milliammeter Current and Resistance:
   • Maximum current through the milliammeter coil:
     $I_m = 10 \, \text{mA} = 0.01 \, \text{A}$ 

     ,

   • Resistance of the milliammeter coil:
     $R_m = 1 \, \Omega$ 

     .

2. Total Current for the Ammeter:
   • The total current to be measured by the modified ammeter:
     $I = 1 \, \text{A}$ 

     .

3. Shunt Current:
   • The shunt carries the remaining current,
     $I_S = I - I_m = 1 \, \text{A} - 0.01 \, \text{A} = 0.99 \, \text{A}$ 

     .

4. Voltage Across Shunt and Milliammeter:
   • The shunt is connected in parallel with the milliammeter, so the voltage across them is the same:
     $$V_m = V_S.$$ 
5. Ohm's Law:
   • Voltage across the milliammeter:
     $$V_m = I_m \cdot R_m = 0.01 \cdot 1 = 0.01 \, \text{V}.$$ 
   • Voltage across the shunt:
     $$V_S = I_S \cdot R_S = 0.01 \, \text{V}.$$ 
6. Solve for Shunt Resistance (
   $R_S$ 

   ):

   • From the voltage equation:
     $$R_S = \frac{V_S}{I_S} = \frac{0.01}{0.99} = \frac{1}{99} \, \Omega.$$ 

Final Answer:

The required shunt resistance is:

$$\boxed{\frac{1}{99} \, \Omega}.$$

To find the resistance

$R$

that must be placed in series with the bulb, let's analyze the problem step by step.

Given:

1. Power of the bulb (
   $P$ 

   ) = 500 W,

2. Voltage rating of the bulb (
   $V_b$ 

   ) = 100 V,

3. Supply voltage (
   $V_s$ 

   ) = 200 V.

Step 1: Resistance of the bulb

The resistance of the bulb (

$R_b$

) can be calculated using the formula:

$$R_b = \frac{V_b^2}{P}.$$

Substitute the values:

$$R_b = \frac{100^2}{500} = \frac{10000}{500} = 20 \, \Omega.$$

Step 2: Total current through the circuit

The bulb is rated to draw 500 W at 100 V. Thus, the current through the bulb is:

$$I = \frac{P}{V_b}.$$

Substitute the values:

$$I = \frac{500}{100} = 5 \, \text{A}.$$

Step 3: Voltage drop across the series resistor

The total supply voltage is 200 V, and the bulb operates at 100 V. Therefore, the voltage drop across the series resistor

$R$

is:

$$V_R = V_s - V_b.$$

Substitute the values:

$$V_R = 200 - 100 = 100 \, \text{V}.$$

Step 4: Resistance of the series resistor

Using Ohm's law, the resistance of the series resistor is:

$$R = \frac{V_R}{I}.$$

Substitute the values:

$$R = \frac{100}{5} = 20 \, \Omega.$$

Final Answer:

The resistance that must be placed in series with the bulb is:

$$\boxed{20 \, \Omega}.$$

To construct an ammeter that can read currents up to

$2.0 \, \text{A}$

, we need to calculate the resistance of the shunt. Here's the step-by-step calculation:

1. Full-scale current through the coil:
   The coil takes $1 \, \text{mA} = 10^{-3} \, \text{A}$ 

   for full-scale deflection.

2. Remaining current through the shunt:
   When the total current is $2.0 \, \text{A}$ 

   , the current through the shunt is:

   $$I_s = I_{\text{total}} - I_c = 2.0 - 10^{-3} = 1.999 \, \text{A}.$$ 

3. Voltage across the coil:
   The resistance of the coil is $25 \, \Omega$ 

   . Using Ohm's law, the voltage across the coil is:

   $$V_c = I_c R_c = (10^{-3})(25) = 0.025 \, \text{V}.$$ 

4. Resistance of the shunt:
   The voltage across the shunt must equal the voltage across the coil:  $$V_s = V_c = 0.025 \, \text{V}.$$Using Ohm's law for the shunt:

    

   $$R_s = \frac{V_s}{I_s} = \frac{0.025}{1.999} \approx 1.25 \times 10^{-2} \, \Omega.$$ 

Thus, the resistance of the shunt is:

$$\boxed{1.25 \times 10^{-2} \, \Omega.}$$

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.

Using the formula for drift velocity \(v_d = \frac{I}{n e A}\), we substitute the given values: \(v_d = \frac{1.5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-7}} \approx 1.1 \times 10^{-3}\text{ m/s} = 1.1\text{ mm s}^{-1}\).