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:

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