A voltmeter consists of a `25Omega` coil connected in series with a `575 Omega` resistor. The coil takes 10 mA for full-scale deflection. What maximum potential difference can be measured on this voltmeter?

Answer

To find the maximum potential difference that can be measured by the voltmeter, we can follow these steps:

### Step 1: Identify the components of the voltmeter
The voltmeter consists of a coil with a resistance of \( R_1 = 25 \, \Omega \) and a series resistor with a resistance of \( R_2 = 575 \, \Omega \).

### Step 2: Calculate the equivalent resistance
Since the coil and the resistor are in series, the total or equivalent resistance \( R_{eq} \) can be calculated using the formula:
\[ R_{eq} = R_1 + R_2 \]
Substituting the values:
\[ R_{eq} = 25 \, \Omega + 575 \, \Omega = 600 \, \Omega \]

### Step 3: Determine the maximum current
The maximum current \( I_{max} \) for full-scale deflection is given as \( 10 \, \text{mA} \). We need to convert this into amperes:
\[ I_{max} = 10 \, \text{mA} = 10 \times 10^{-3} \, \text{A} = 0.01 \, \text{A} \]

### Step 4: Use Ohm's Law to find the maximum potential difference
According to Ohm's Law, the potential difference \( V \) across a resistor is given by:
\[ V = I \times R \]
Thus, the maximum potential difference \( V_{max} \) that can be measured by the voltmeter is:
\[ V_{max} = I_{max} \times R_{eq} \]
Substituting the values we found:
\[ V_{max} = 0.01 \, \text{A} \times 600 \, \Omega \]

### Step 5: Calculate \( V_{max} \)
Calculating the above expression:
\[ V_{max} = 0.01 \times 600 = 6 \, \text{V} \]

### Final Answer
The maximum potential difference that can be measured on this voltmeter is \( 6 \, \text{V} \).

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