A cell of emf `3.4 V` and internal resistance `3 Omega` is connected to an ammeter having resistance `2Omega` and to an external resistance of `100Omega`. When a voltmeter is connected across the `100 Omega` resistance, the ammeter reading is `0.04 A`. Find the voltage reading by the voltmeter and its resistance. Had the voltmeter been an ideal one what would have been its reading?

Explanation

To solve the given problem, we will follow these steps:

### Step 1: Understand the Circuit Configuration
We have a circuit consisting of:
- A cell with an EMF of \(3.4 \, V\) and an internal resistance \(r = 3 \, \Omega\).
- An ammeter with a resistance \(R_a = 2 \, \Omega\).
- An external resistor \(R_e = 100 \, \Omega\).
- A voltmeter connected across the \(100 \, \Omega\) resistor.

### Step 2: Calculate the Total Resistance in the Circuit
The total resistance in the circuit is the sum of the internal resistance of the cell, the resistance of the ammeter, and the equivalent resistance of the external resistor and the voltmeter in parallel.

Let \(R_v\) be the resistance of the voltmeter. The equivalent resistance \(R_{AB}\) across the \(100 \, \Omega\) resistor and the voltmeter can be calculated using the formula for parallel resistors:
\[ R_{AB} = \frac{100 \times R_v}{100 + R_v} \]

Thus, the total resistance \(R_{total}\) in the circuit is:
\[ R_{total} = r + R_a + R_{AB} = 3 + 2 + \frac{100 \times R_v}{100 + R_v} \]

### Step 3: Use the Current Reading to Find \(R_v\)
The current \(I\) in the circuit is given as \(0.04 \, A\). According to Ohm's Law:
\[ I = \frac{V}{R_{total}} \]
Substituting the values:
\[ 0.04 = \frac{3.4}{3 + 2 + \frac{100 \times R_v}{100 + R_v}} \]

### Step 4: Solve for \(R_v\)
Rearranging the equation:
\[ 0.04 \left(3 + 2 + \frac{100 \times R_v}{100 + R_v}\right) = 3.4 \]
\[ 0.04 \left(5 + \frac{100 \times R_v}{100 + R_v}\right) = 3.4 \]
Expanding:
\[ 0.2 + \frac{4R_v}{100 + R_v} = 3.4 \]
Subtracting \(0.2\) from both sides:
\[ \frac{4R_v}{100 + R_v} = 3.2 \]
Cross-multiplying:
\[ 4R_v = 3.2(100 + R_v) \]
Expanding:
\[ 4R_v = 320 + 3.2R_v \]
Rearranging:
\[ 4R_v - 3.2R_v = 320 \]
\[ 0.8R_v = 320 \]
Thus,
\[ R_v = \frac{320}{0.8} = 400 \, \Omega \]

### Step 5: Calculate the Voltage Reading by the Voltmeter
Now, we can find the voltage across the \(100 \, \Omega\) resistor (which is the reading of the voltmeter):
\[ R_{AB} = \frac{100 \times 400}{100 + 400} = \frac{40000}{500} = 80 \, \Omega \]
Now, the total resistance becomes:
\[ R_{total} = 3 + 2 + 80 = 85 \, \Omega \]
The current through the circuit is still \(0.04 \, A\), so the voltage across \(R_{AB}\) is:
\[ V_{AB} = I \times R_{AB} = 0.04 \times 80 = 3.2 \, V \]

### Step 6: Ideal Voltmeter Reading
If the voltmeter were ideal (i.e., \(R_v = 0\)), the total resistance would be:
\[ R_{total} = 3 + 2 + 100 = 105 \, \Omega \]
The current would then be:
\[ I = \frac{3.4}{105} \approx 0.0324 \, A \]
The voltage across the \(100 \, \Omega\) resistor would be:
\[ V_{ideal} = I \times 100 = 0.0324 \times 100 \approx 3.24 \, V \]

### Final Answers
- The voltage reading by the voltmeter is \(3.2 \, V\).
- The resistance of the voltmeter is \(400 \, \Omega\).
- If the voltmeter were ideal, its reading would be approximately \(3.24 \, V\).