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A 5 A, 110 V electrodynamic type wattmeter has a scale having 110 divisions. Its pressure coil is fed by a voltage of [110 √2 cos314) + √2sin(942t) V and its current coil carries a current of [5√ 2cos(314t + 60) + 2√ 2sin (628t + 90) + √2cos(642t + 90)] A. The needle of the wattmeter will move to1. 110 divisions2. 50 divisions3. 54 divisions4. 55 divisions |
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Answer» Correct Answer - Option 4 : 55 divisions The average power read by the wattmeter meter, is \( = { \vee _{rms\;}}{I_{rms}}cos\phi = \frac{{110\;\sqrt 2 }}{{\sqrt 2 }} - \frac{{5\sqrt 2 }}{{\sqrt 2 }}{\rm{cos}}\left( {60°} \right)\) \(= 110 \times 5 \times \frac{1}{2} = \frac{{550}}{2}\omega \) Full scale reading of the meter = 110 × 5 = 500 ω 550 ω = 110 divisions \(\frac{{550}}{2}\omega = \frac{{110}}{2} \Rightarrow 55\;divisions\) |
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