Physics, asked by shubhankar5264, 11 months ago

V=5cos(wt-π/6)and I= 10sin(wt). Then what is the average power dissipated in the circuit

Answers

Answered by jaisngh720
0

A/C to question,

i = 5cosωt

i = 5sin(ωt + π/2) [ we know , sin(90+A) = cosA ]

V = 200sinωt

V = 200sin(ωt + 0)

hence, phase difference between I and V is

Φ = π/2 - 0 = π/2

now, Power loss in circuit = Vi.cosΦ

where V and I are sinusoids rms values, Φ is the phase difference between Voltage and current.

because Φ = π/2 => cosΦ = 0

so, Power loss in circuit = 0

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