A voltmeter has a resistance of `50 Omega` is connected across a cell of e.m.f. 2V and internal resistance `10 Omega`. The reading of voltmeter is

Options

A. 1.667V
B. 16.7V
C. 167V
D. 0.167V

Explanation

To solve the problem, we need to find the reading of the voltmeter when it is connected across a cell with a given e.m.f. and internal resistance. Here are the steps to arrive at the solution:

### Step 1: Identify the given values
- E.M.F. of the cell (V) = 2 V
- Internal resistance of the cell (r) = 10 Ω
- Resistance of the voltmeter (R_v) = 50 Ω

### Step 2: Calculate the equivalent resistance in the circuit
The total resistance in the circuit (R_eq) is the sum of the internal resistance of the cell and the resistance of the voltmeter since they are in series:
\[ R_{eq} = r + R_v = 10 \, \Omega + 50 \, \Omega = 60 \, \Omega \]

### Step 3: Calculate the current in the circuit using Ohm's Law
Using Ohm's Law, we can find the current (I) flowing through the circuit:
\[ V = I \cdot R_{eq} \]
Rearranging gives:
\[ I = \frac{V}{R_{eq}} = \frac{2 \, V}{60 \, \Omega} = \frac{1}{30} \, A \]

### Step 4: Calculate the potential drop across the voltmeter
The reading of the voltmeter (V_v) can be calculated by finding the potential drop across the voltmeter:
\[ V_v = I \cdot R_v = \left(\frac{1}{30} \, A\right) \cdot (50 \, \Omega) = \frac{50}{30} \, V = \frac{5}{3} \, V \]

### Step 5: Convert the result to decimal form
To express the result in decimal form:
\[ \frac{5}{3} \approx 1.67 \, V \]

### Conclusion
The reading of the voltmeter when connected across the cell is approximately **1.67 V**.

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