The resistor of resistance ' R ' is connected to 25 V supply and heat produced in it is 25 J / sec. The value of R is

Options

A. `225Omega`
B. `1Omega`
C. `25Omega`
D. `50Omega`

Answer

To solve the problem, we need to find the resistance \( R \) given that a resistor connected to a 25 V supply produces heat at a rate of 25 J/s (which is equivalent to 25 W).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the power \( P \) dissipated in a resistor due to heat is given by Joule's law, which states:
\[ P = I^2 R \]
where \( P \) is the power in watts (W), \( I \) is the current in amperes (A), and \( R \) is the resistance in ohms (Ω).

2. **Given Values**:
- Voltage \( V = 25 \, \text{V} \)
- Power \( P = 25 \, \text{W} \)

3. **Using Ohm's Law**:
According to Ohm's law, the relationship between voltage, current, and resistance is given by:
\[ V = I R \]
From this, we can express current \( I \) in terms of voltage and resistance:
\[ I = \frac{V}{R} \]

4. **Substituting into the Power Equation**:
Now, substitute \( I \) from Ohm's law into the power equation:
\[ P = \left(\frac{V}{R}\right)^2 R \]
Simplifying this gives:
\[ P = \frac{V^2}{R} \]

5. **Rearranging to Find Resistance**:
We can rearrange the equation to solve for \( R \):
\[ R = \frac{V^2}{P} \]

6. **Substituting the Known Values**:
Now substitute the known values of \( V \) and \( P \):
\[ R = \frac{25^2}{25} \]
\[ R = \frac{625}{25} \]
\[ R = 25 \, \Omega \]

7. **Conclusion**:
Therefore, the value of the resistance \( R \) is \( 25 \, \Omega \).