The current i passed in any instrument in an A.C. circuit is i=2sinωt amp and potential difference applied is given by V=5cosωt volt. Power loss in the instrument is:10 watt5 wattzero watt20 watt
Question
The current i passed in any instrument in an A.C. circuit is i=2sinωt amp and potential difference applied is given by V=5cosωt volt. Power loss in the instrument is:10 watt5 wattzero watt20 watt
Solution
The power loss in an AC circuit can be calculated using the formula P = VIcosφ, where V is the voltage, I is the current, and φ is the phase difference between the voltage and the current.
In this case, the current i = 2sinωt and the voltage V = 5cosωt.
The phase difference φ can be determined by comparing the functions of current and voltage. Here, the current is a sine function and the voltage is a cosine function, which means they are phase-shifted by 90 degrees or π/2 radians.
The cosine of π/2 is 0, so the power loss P = VIcosφ = 250 = 0 watt.
So, the power loss in the instrument is zero watt.
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