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In the figure Vs = 10 cos t = 10 0°. The current drawn from Vs is given by the phasor—
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2 – 45° 5 - 2 2 45°
- 2 0°
- 2 90º
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Correct Option: B
For the given circuit 
| Z (s) = 5 || | = | 1 / 5(s) | 5 + (5 / s) |
| Z (s) = | = | 5s + 5 | s + 1 |
| I (s) = | = | Z (s) | 5 / s + 1 |
or I (s) = 2 ∠ 0° (jω + 1)
{ ∴ 10 cos t = 10 ∠ 0° ∵ ω = 1 }
= 2 ∠ 0° √2. ∠ 45°
= 2 √2 ∠ 45°
