To measure a potential difference across a resistor of resistance RS, a voltmeter of resistance Ry is used. To measure the potential with a minimum accuracy of 95%, then

Options

A. `R_(v) = 5R`
B. `R_(v) = 15 R`
C. `R_(v) = 10 R`
D. `R_(v),ge 19 R `

Answer

To solve the problem of measuring the potential difference across a resistor \( R_S \) using a voltmeter of resistance \( R_Y \) with a minimum accuracy of 95%, we can follow these steps:

### Step 1: Understand the Circuit Configuration
The voltmeter is connected in parallel with the resistor \( R_S \). This means that the voltage across the resistor \( R_S \) (denoted as \( V_0 \)) is the same as the voltage across the voltmeter (denoted as \( V \)).

### Step 2: Write the Expression for Voltage
The voltage across the resistor \( R_S \) when a current \( I \) flows through it is given by:
\[ V_0 = I \cdot R_S \]

### Step 3: Determine the Voltage Reading of the Voltmeter
Since the voltmeter is in parallel with \( R_S \), the effective resistance \( R_{eq} \) of the parallel combination of \( R_S \) and \( R_Y \) can be calculated using the formula:
\[ R_{eq} = \frac{R_S \cdot R_Y}{R_S + R_Y} \]
The voltage reading \( V \) across the voltmeter can be expressed as:
\[ V = I \cdot R_{eq} = I \cdot \frac{R_S \cdot R_Y}{R_S + R_Y} \]

### Step 4: Set Up the Accuracy Condition
To achieve a minimum accuracy of 95%, the reading \( V \) must be at least 95% of the actual voltage \( V_0 \):
\[ V \geq 0.95 \cdot V_0 \]
Substituting the expressions for \( V \) and \( V_0 \):
\[ I \cdot \frac{R_S \cdot R_Y}{R_S + R_Y} \geq 0.95 \cdot (I \cdot R_S) \]
We can cancel \( I \) from both sides (assuming \( I \neq 0 \)):
\[ \frac{R_S \cdot R_Y}{R_S + R_Y} \geq 0.95 \cdot R_S \]

### Step 5: Rearranging the Inequality
Rearranging the inequality gives:
\[ R_S \cdot R_Y \geq 0.95 \cdot R_S \cdot (R_S + R_Y) \]
Expanding the right side:
\[ R_S \cdot R_Y \geq 0.95 \cdot R_S^2 + 0.95 \cdot R_S \cdot R_Y \]
Rearranging terms:
\[ R_S \cdot R_Y - 0.95 \cdot R_S \cdot R_Y \geq 0.95 \cdot R_S^2 \]
Factoring out \( R_Y \):
\[ (1 - 0.95) R_S \cdot R_Y \geq 0.95 \cdot R_S^2 \]
This simplifies to:
\[ 0.05 \cdot R_S \cdot R_Y \geq 0.95 \cdot R_S^2 \]
Dividing both sides by \( 0.05 \cdot R_S \) (assuming \( R_S \neq 0 \)):
\[ R_Y \geq 19 \cdot R_S \]

### Conclusion
Thus, for the voltmeter resistance \( R_Y \) to ensure a minimum accuracy of 95%, it must satisfy the condition:
\[ R_Y \geq 19 \cdot R_S \]