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
We have a resistor \( R_S \) and a voltmeter with resistance \( R_Y \) connected in parallel to measure the voltage across \( R_S \). According to Ohm's Law, the voltage across the resistor can be expressed as:
\[ V_0 = I \cdot R_S \]
where \( I \) is the current flowing through the circuit.

### Step 2: Voltage Across the Voltmeter
The voltmeter reading \( V \) can be expressed as the voltage across the parallel combination of \( R_S \) and \( R_Y \):
\[ V = I \cdot \frac{R_S \cdot R_Y}{R_S + R_Y} \]

### Step 3: Set Up the Accuracy Condition
To achieve a minimum accuracy of 95%, we need the voltmeter reading \( V \) to be at least 95% of the actual voltage \( V_0 \):
\[ V > 0.95 \cdot V_0 \]
Substituting the expressions for \( V \) and \( V_0 \):
\[ I \cdot \frac{R_S \cdot R_Y}{R_S + R_Y} > 0.95 \cdot (I \cdot R_S) \]

### Step 4: Simplify the Inequality
We can cancel \( I \) from both sides (assuming \( I \neq 0 \)):
\[ \frac{R_S \cdot R_Y}{R_S + R_Y} > 0.95 \cdot R_S \]

### Step 5: Rearranging the Terms
Rearranging gives:
\[ R_S \cdot R_Y > 0.95 \cdot R_S \cdot (R_S + R_Y) \]
Expanding the right side:
\[ R_S \cdot R_Y > 0.95 \cdot R_S^2 + 0.95 \cdot R_S \cdot R_Y \]

### Step 6: Isolate \( R_Y \)
Rearranging the terms leads to:
\[ R_S \cdot R_Y - 0.95 \cdot R_S \cdot R_Y > 0.95 \cdot R_S^2 \]
Factoring out \( R_Y \):
\[ R_Y (1 - 0.95) > 0.95 \cdot R_S \]
This simplifies to:
\[ 0.05 \cdot R_Y > 0.95 \cdot R_S \]

### Step 7: Solve for \( R_Y \)
Dividing both sides by 0.05 gives:
\[ R_Y > \frac{0.95}{0.05} \cdot R_S \]
Calculating the right side:
\[ R_Y > 19 \cdot R_S \]

### Conclusion
Thus, to measure the potential difference with a minimum accuracy of 95%, the resistance of the voltmeter \( R_Y \) must be greater than \( 19 \cdot R_S \).

### Final Answer
\[ R_Y \geq 19 \cdot R_S \]

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