An ammeter and a voltmeter of resistance `R` connected in seires to an electric cell of negligible internal resistance. Their readings are `A` and `V` respecitvely. If another resistance `R` is connected in parallel with the voltmeter

Options

A. both A and V will increase
B. both A and V will decrease
C. A will decrease and V will increase
D. A will increase and V will decrease

Answer

To solve the problem step-by-step, we will analyze the circuit with the ammeter, voltmeter, and the additional resistance connected in parallel with the voltmeter.

### Step 1: Understand the initial circuit
Initially, we have an ammeter (A) and a voltmeter (V) connected in series to an electric cell with negligible internal resistance. The resistance of both the ammeter and voltmeter is given as \( R \).

### Step 2: Determine the current through the circuit
The reading of the ammeter is \( A \). Therefore, the current flowing through the circuit is:
\[ I = A \]

### Step 3: Calculate the voltage across the voltmeter
The voltage across the voltmeter can be expressed using Ohm's law:
\[ V = I \times R \]
Substituting the value of \( I \):
\[ V = A \times R \quad \text{(Equation 1)} \]

### Step 4: Analyze the total resistance in the circuit
The total resistance in the circuit (since the ammeter and voltmeter are in series) is:
\[ R_{\text{total}} = R + R = 2R \]
The emf of the cell can be expressed as:
\[ V' = I \times R_{\text{total}} = A \times 2R \quad \text{(Equation 2)} \]

### Step 5: Introduce the new resistance in parallel with the voltmeter
When another resistance \( R \) is connected in parallel with the voltmeter, the effective resistance \( R_{\text{eff}} \) of the voltmeter and the new resistance in parallel can be calculated as:
\[ R_{\text{eff}} = \frac{R \times R}{R + R} = \frac{R^2}{2R} = \frac{R}{2} \]

### Step 6: Calculate the new total resistance
Now, the new total resistance in the circuit becomes:
\[ R_{\text{new}} = R + R_{\text{eff}} = R + \frac{R}{2} = \frac{3R}{2} \]

### Step 7: Calculate the new current in the circuit
Using Ohm's law, the new current \( I' \) in the circuit can be calculated as:
\[ I' = \frac{V'}{R_{\text{new}}} = \frac{V'}{\frac{3R}{2}} = \frac{2V'}{3R} \]
Substituting \( V' = 2AR \) from Equation 2:
\[ I' = \frac{2 \times 2AR}{3R} = \frac{4A}{3} \quad \text{(Equation 3)} \]

### Step 8: Calculate the new reading of the voltmeter
The new reading of the voltmeter \( G \) can be calculated using the effective resistance:
\[ G = I' \times R_{\text{eff}} = \left(\frac{4A}{3}\right) \times \frac{R}{2} = \frac{4AR}{6} = \frac{2V}{3} \quad \text{(Equation 4)} \]

### Step 9: Conclusion
From the calculations, we have:
- The new current \( I' = \frac{4A}{3} \) (which is greater than \( A \))
- The new voltmeter reading \( G = \frac{2V}{3} \) (which is less than \( V \))

### Summary of Results
- The ammeter reading increases: \( A' = \frac{4A}{3} \)
- The voltmeter reading decreases: \( G = \frac{2V}{3} \)

### Final Answer
The correct conclusion is that the ammeter reading increases while the voltmeter reading decreases.