1: Analyzing the Circuit
The given circuit consists of resistors in series and parallel. First, simplify the resistances step by step. 1. Combine the 10\(\mu\) and 10\(\mu\) resistors in parallel between points B and C: \[ R_{\text{BC}} = \frac{1}{\frac{1}{10} + \frac{1}{10}} = 5 \] 2. Combine the 20\(\mu\) resistors in parallel between points C and D: \[ R_{\text{CD}} = \frac{1}{\frac{1}{20} + \frac{1}{20}} = 10 \] 3. Simplify the overall network of resistors to find the effective resistance between points A and M: \[ R_{\text{eff}} = 12 \]
2: Power Supplied by the Battery
The power supplied by the battery is given by: \[ P = \frac{V^2}{R_{\text{eff}}} = \frac{6^2}{12} = 3W \] Thus, the power supplied by the battery is \( P = 3W \).
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 |