If alternating current of rms value ‘a’ flows through resistance R then power loss in resistance is-

Options

A. zero
B. `a^(2)`R
C. `(a^(2)R)/(2)`
D. `2a^(2)R `

Answer

To find the power loss in a resistance when an alternating current of RMS value 'a' flows through it, we can follow these steps:

### Step-by-step Solution:

1. **Understanding RMS Current**:
The RMS (Root Mean Square) value of the alternating current is given as 'a'. This value represents the effective value of the current that would produce the same power as a DC current.

2. **Power Formula**:
The average power (P) dissipated in a resistor when an alternating current flows through it can be calculated using the formula:
\[ P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos(\phi) \]
where \( V_{\text{rms}} \) is the RMS voltage, \( I_{\text{rms}} \) is the RMS current, and \( \phi \) is the phase difference between the current and voltage.

3. **For a Purely Resistive Circuit**:
In a purely resistive circuit, the phase difference \( \phi \) is 0 degrees. Therefore, \( \cos(\phi) = \cos(0) = 1 \).

4. **Relating Voltage and Current**:
For a resistive load, Ohm's law states that:
\[ V_{\text{rms}} = I_{\text{rms}} \cdot R \]
Given that \( I_{\text{rms}} = a \), we can express the RMS voltage as:
\[ V_{\text{rms}} = a \cdot R \]

5. **Substituting into Power Formula**:
Now, substituting \( V_{\text{rms}} \) into the power formula:
\[ P = (a \cdot R) \cdot a \cdot 1 \]
Simplifying this gives:
\[ P = a^2 \cdot R \]

6. **Conclusion**:
Thus, the power loss in the resistance when an alternating current of RMS value 'a' flows through it is:
\[ P = a^2 R \]

### Final Answer:
The power loss in resistance is \( P = a^2 R \).

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