What reading would you expact of a square-wave current, suitching rapodly between +0.5 A and -0.5 A, when passed through an ac ammeter?

Options

A. 0
B. 0.5A
C. 0.25A
D. 1.0A

Explanation

To solve the problem of determining the reading of a square-wave current switching rapidly between +0.5 A and -0.5 A when passed through an AC ammeter, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the AC Ammeter**:
- An AC ammeter measures the root mean square (RMS) value of the alternating current.

2. **Defining the Square Wave Current**:
- The square wave current alternates between +0.5 A and -0.5 A. This means that for half the time (T/2), the current is +0.5 A, and for the other half (T/2), it is -0.5 A.

3. **Calculating the RMS Value**:
- The RMS value of a current is defined as the square root of the average of the squares of the instantaneous values over one complete cycle.
- For a square wave, the formula for RMS current (I_rms) can be simplified as:
\[ I_{rms} = \sqrt{\frac{1}{T} \int_0^T I^2(t) dt} \]

4. **Setting Up the Integral**:
- We can break the integral into two parts:
- From 0 to T/2, where I = +0.5 A
- From T/2 to T, where I = -0.5 A
- Thus, the integral becomes:
\[ I_{rms} = \sqrt{\frac{1}{T} \left( \int_0^{T/2} (0.5)^2 dt + \int_{T/2}^{T} (-0.5)^2 dt \right)} \]

5. **Calculating Each Integral**:
- For the first part (0 to T/2):
\[ \int_0^{T/2} (0.5)^2 dt = (0.5)^2 \cdot \frac{T}{2} = 0.25 \cdot \frac{T}{2} = 0.125 T \]
- For the second part (T/2 to T):
\[ \int_{T/2}^{T} (-0.5)^2 dt = (-0.5)^2 \cdot \frac{T}{2} = 0.25 \cdot \frac{T}{2} = 0.125 T \]

6. **Combining the Results**:
- Now, we add the two results:
\[ \int_0^T I^2(t) dt = 0.125 T + 0.125 T = 0.25 T \]

7. **Final Calculation of RMS Current**:
- Plugging this back into the RMS formula:
\[ I_{rms} = \sqrt{\frac{1}{T} \cdot 0.25 T} = \sqrt{0.25} = 0.5 \text{ A} \]

8. **Conclusion**:
- The reading on the AC ammeter will be **0.5 A**.