In an AC circuit, the voltage applied is E = E0 sin ω t . The resulting current in the circuit is I = I0 sin (ω t - π/2). The power consumption in the circuit is given by

Question

In an AC circuit, the voltage applied is E = E0 sin ω t . The resulting current in the circuit is I = I0 sin (ω t - π/2). The power consumption in the circuit is given by (A) P = 0 (B) P = (E0 I0/2) (C) P = √2 E0 I0 (D) P = (E0 I0/√2)

Tardigrade Admin · Accepted Answer

Pav = EvIv cos φ Here φ = Phase difference = (π/2) ∴ Pav = 0

Q. In an $A C$ circuit, the voltage applied is $E = E_{0} s in ω t$ . The resulting current in the circuit is $I = I_{0} s in (ω t - π /2)$ . The power consumption in the circuit is given by

$P = 0$

$P = \frac{E _{0} I _{0}}{2}$

$P = 2 E_{0} I_{0}$

$P = \frac{E _{0} I _{0}}{2}$

$P_{a v} = E_{v} I_{v} cos ϕ$
Here $ϕ =$ Phase difference $= \frac{π}{2}$
$∴ P_{a v} = 0$

Q. In an AC circuit, the voltage applied is E=E0​sinωt . The resulting current in the circuit is I=I0​sin(ωt−π/2). The power consumption in the circuit is given by

P=0

P=2E0​I0​​

P=2​E0​I0​

P=2​E0​I0​​

Pav​=Ev​Iv​cosϕ Here ϕ=Phase difference =2π​ ∴Pav​=0

Q. In an $A C$ circuit, the voltage applied is $E = E_{0} s in ω t$ . The resulting current in the circuit is $I = I_{0} s in (ω t - π /2)$ . The power consumption in the circuit is given by

$P = 0$

$P = \frac{E _{0} I _{0}}{2}$

$P = 2 E_{0} I_{0}$

$P = \frac{E _{0} I _{0}}{2}$

$P_{a v} = E_{v} I_{v} cos ϕ$
Here $ϕ =$ Phase difference $= \frac{π}{2}$
$∴ P_{a v} = 0$