Question

The current coil of a 200 V, 5 A, electrodynamoter type LPF wattmeter carries a current of $\sqrt{2}cos\left(100\pi t\right)A.$ The voltage across the pressure coil is $\sqrt{2}sin\left(100\pi t\right)V.$ The meter will indicate

• A. 0 W
• B. 100 W
• C. 200 W
• D. 400 W

Answer 1

The correct option is A 0 W

$Readingofwattmeter=V_{PC}{I}_{CC}cos\varphi$

$where,fisanglebetween{V}_{PC}\text{and}{I}_{CC}.$

Now given,

$I_{CC}=\sqrt{2}cos\left(100\pi t\right)A$

$=\sqrt{2}sin(100\pi t+{90}^{\circ })A$

$V_{PC}=\sqrt{2}sin\left(100\pi t\right)V$

$P=VIcos\left({90}^{\circ }\right)$

$=0W$