Alternating CurrentMedium
Question
The potential difference V across and the current I flowing through an instrument in an AC circuit are given by :
V = 5 cos ω t volt
I = 2 sin ω t volt
The power dissipated in the instrument is :
V = 5 cos ω t volt
I = 2 sin ω t volt
The power dissipated in the instrument is :
Options
A.zero
B.5 watt
C.10 watt
D.2.5 watt
Solution
Pav = vrms Irms cos φ
Here φ = 90o so Pav = 0
φ = 90o Pav = 0
Here φ = 90o so Pav = 0
φ = 90o Pav = 0
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