In an AC circuit V and I are given below, then find the power dissipated in the circuit V=50 sin (50 t) ∨ I=50 sin (50 t (π/3)) mA

Question

In an AC circuit V and I are given below, then find the power dissipated in the circuit V=50 sin (50 t) ∨ I=50 sin (50 t (π/3)) mA (A) 0.625 W (B) 1.25 W (C) 2.50 W (D) 5.0 W

Tardigrade Admin · Accepted Answer

As we know that, the power dissipated in an AC
circuit is

P = \frac{E _{0} i ˙ c o s ϕ}{2}

Here,

E_{0} = 50 V

i_{0} = 50 m A = 50 \times 1 0^{- 3} A and ϕ = \frac{π}{3}

(i.e. cos \frac{π}{3} = \frac{1}{2})

∴ P = \frac{50 \times 50 \times 1 0 ^{- 3} \times 1}{2 \times 2} = 0.625 W

Q. In an AC circuit V and I are given below, then find the power dissipated in the circuit V=50sin(50t)∨I=50sin(50t3π​)mA

0.625 W

1.25 W

2.50 W

5.0 W

As we know that, the power dissipated in an AC circuit is P=2E0​i˙cosϕ​ Here, E0​=50V i0​=50mA=50×10−3A and ϕ=3π​ ( i.e. cos3π​=21​) ∴P=2×250×50×10−3×1​=0.625W

Q. In an AC circuit $V$ and $I$ are given below, then find the power dissipated in the circuit $V = 50 sin (50 t) \lor I = 50 sin (50 t \frac{π}{3}) m A$

As we know that, the power dissipated in an AC
circuit is $P = \frac{E _{0} i ˙ c o s ϕ}{2}$
Here, $E_{0} = 50 V$
$i_{0} = 50 m A = 50 \times 1 0^{- 3} A and ϕ = \frac{π}{3}$
$(i.e. cos \frac{π}{3} = \frac{1}{2})$
$∴ P = \frac{50 \times 50 \times 1 0 ^{- 3} \times 1}{2 \times 2} = 0.625 W$