In an AC circuit the potential difference V and current I are given respectively by V=100 sin (100t)volt and I=100 sin ( 100t+(π /3) )mA The power dissipated in the circuit will be

Question

In an AC circuit the potential difference V and current I are given respectively by V=100 sin (100t)volt and I=100 sin ( 100t+(π /3) )mA The power dissipated in the circuit will be (A)  104W  (B)  10W  (C)  2.5 text W  (D)  5W

Tardigrade Admin · Accepted Answer

Voltage amplitude V0=100 V, current amplitude I0=100 mA=100× 10-3A Phase difference φ =(π /3)=60° Power dissipated p=(V0I0/2) cos φ =(100× 100× 10-3/2)× cos 60° =2.5 W

Q. In an AC circuit the potential difference V and current I are given respectively by $V = 100 sin (100 t) v o lt$ and $I = 100 sin (100 t + \frac{π}{3}) m A$ The power dissipated in the circuit will be

$10^{4} W$

$10 W$

$2.5 W$

$5 W$

Voltage amplitude $V_{0} = 100 V,$ current amplitude $I_{0} = 100 m A = 100 \times 10^{- 3} A$ Phase difference $ϕ = \frac{π}{3} = 60^{\circ}$ Power dissipated $p = \frac{V _{0} I _{0}}{2} cos ϕ$ $= \frac{100 \times 100 \times 10 ^{- 3}}{2} \times cos 60^{\circ}$ $= 2.5 W$

Q. In an AC circuit the potential difference V and current I are given respectively by V=100sin(100t)volt and I=100sin(100t+3π​)mA The power dissipated in the circuit will be

104W

10W

2.5W

5W