In an AC circuit- EMF applied is -5sin- t due to which a current i-3cos- t flows in the circuit- The average power in watt dissipated in the circuit will be

Given, ℰ=5sinω t and nbsp; i=3cosω t=3sin(π2 ω t) [∵cosθ=sin(π2 θ) So, the phase difference, ϕ=π /2. Average power dissipated, P_avg=ℰ_rmsI_rmscosϕ ...

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Standard XII

Physics

Zener Diode

In an AC circ...

Question

In an $AC$ circuit, $EMF$ applied is $E = 5 sin ω t$ due to which a current $i = 3 cos ω t$ flows in the circuit. The average power $(in watt)$ dissipated in the circuit will be

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Solution

Given,
$E = 5 sin ω t$ and

$i = 3 cos ω t = 3 sin (\frac{π}{2} - ω t) [∵ cos θ = sin (\frac{π}{2} - θ)$

So, the phase difference, $ϕ = π / 2$ .

Average power dissipated,

$P_{a v g} = E_{r m s} I_{r m s} cos ϕ = 0 [∵ ϕ = \frac{π}{2}]$

Correct Answer: $0$

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In an AC circuit, EMF applied is E=5sinωt due to which a current i=3cosωt flows in the circuit. The average power (in watt) dissipated in the circuit will be

Given, E=5sinωt and i=3cosωt=3sin(π2−ωt) [∵cosθ=sin(π2−θ) So, the phase difference, ϕ=π/2. Average power dissipated, Pavg=ErmsIrmscosϕ=0 [∵ϕ=π2] Correct Answer: 0

In an $AC$ circuit, $EMF$ applied is $E = 5 sin ω t$ due to which a current $i = 3 cos ω t$ flows in the circuit. The average power $(in watt)$ dissipated in the circuit will be

Given,
$E = 5 sin ω t$ and

$i = 3 cos ω t = 3 sin (\frac{π}{2} - ω t) [∵ cos θ = sin (\frac{π}{2} - θ)$

So, the phase difference, $ϕ = π / 2$ .

Average power dissipated,

$P_{a v g} = E_{r m s} I_{r m s} cos ϕ = 0 [∵ ϕ = \frac{π}{2}]$

Correct Answer: $0$