Understanding Rectifier Frequency Response
- The given AC voltage is: \[ V = 0.5 \sin(100\pi t) \] - The general form of an AC signal is: \[ V = V_0 \sin(2\pi f t) \] Comparing, we get: \[ 2\pi f = 100\pi \] \[ f = \frac{100\pi}{2\pi} = 50 \text{ Hz} \]
Half-Wave Rectifier Output Frequency:
- A half-wave rectifier allows only one half-cycle of the input AC voltage.
- The output voltage still follows the same fundamental frequency as the input, i.e., 50 Hz. Full
-Wave Rectifier Output Frequency:
- A full-wave rectifier inverts the negative half-cycles, making the output frequency twice the input frequency.
- Thus, the frequency of the output becomes: \[ 2 \times 50 = 100 \text{ Hz} \] Thus, the correct answer is 50 Hz for the half-wave rectifier and 100 Hz for the full-wave rectifier.
The alternating current \( I \) in an inductor is observed to vary with time \( t \) as shown in the graph for a cycle.
Which one of the following graphs is the correct representation of wave form of voltage \( V \) with time \( t \)?}
Commodities | 2009-10 | 2010-11 | 2015-16 | 2016-17 |
---|---|---|---|---|
Agriculture and allied products | 10.0 | 9.9 | 12.6 | 12.3 |
Ore and minerals | 4.9 | 4.0 | 1.6 | 1.9 |
Manufactured goods | 67.4 | 68.0 | 72.9 | 73.6 |
Crude and petroleum products | 16.2 | 16.8 | 11.9 | 11.7 |
Other commodities | 1.5 | 1.2 | 1.1 | 0.5 |
Categories of Reporting Area | As a percentage of total cultivable land (1950-51) | As a percentage of total cultivable land (2014-15) | Area (1950-51) | Area (2014-15) |
---|---|---|---|---|
Culturable waste land | 8.0 | 4.0 | 13.4 | 6.8 |
Fallow other than current fallow | 6.1 | 3.6 | 10.2 | 6.2 |
Current fallow | 3.7 | 4.9 | 6.2 | 8.4 |
Net area sown | 41.7 | 45.5 | 70.0 | 78.4 |
Total Cultivable Land | 59.5 | 58.0 | 100.00 | 100.00 |