A ammeter of resistance `0.2Omega` and range 10mA is to be used to read potential difference upto 1V. What is to be done?

Options

A. `96Omega` in series
B. `92Omega` in parallel
C. `99.8Omega` in series
D. `90Omega` in parallel

Explanation

To solve the problem of using an ammeter with a resistance of 0.2 Ω and a range of 10 mA to measure a potential difference of up to 1 V, we can follow these steps:

### Step 1: Understand the Ammeter Specifications
The ammeter has:
- Resistance (R_a) = 0.2 Ω
- Maximum current (I_max) = 10 mA = 10 × 10^(-3) A = 0.01 A

### Step 2: Calculate the Voltage Drop Across the Ammeter
The voltage drop (V_a) across the ammeter when it is at its maximum current can be calculated using Ohm's law:
\[ V_a = I \times R_a \]
Substituting the values:
\[ V_a = 0.01 \, \text{A} \times 0.2 \, \Omega = 0.002 \, \text{V} \]
This means the ammeter can only measure a voltage drop of 2 mV at its maximum current.

### Step 3: Determine the Required Voltage Measurement
We need to measure a potential difference of up to 1 V. To achieve this, we can add an additional resistance (R_g) in series with the ammeter.

### Step 4: Set Up the Equation for Total Voltage
The total voltage (V_total) across the series combination of the ammeter and the additional resistance can be expressed as:
\[ V_{total} = I \times (R_a + R_g) \]
Where:
- V_total = 1 V (the maximum voltage we want to measure)
- I = 10 mA = 0.01 A

### Step 5: Rearrange the Equation to Find R_g
Substituting the known values into the equation:
\[ 1 = 0.01 \times (0.2 + R_g) \]
Now, solve for R_g:
\[ 1 = 0.002 + 0.01 R_g \]
\[ 1 - 0.002 = 0.01 R_g \]
\[ 0.998 = 0.01 R_g \]
\[ R_g = \frac{0.998}{0.01} = 99.8 \, \Omega \]

### Step 6: Conclusion
To enable the ammeter to measure a potential difference of up to 1 V, an additional resistance of approximately 99.8 Ω must be connected in series with the ammeter.