Question

In an AC circuit, V and I are given by V = 50sin50t and I =$100\sin (50t + \dfrac{\pi }{3})mA$. The power dissipated in the circuit.
A. 2.5W
B. 1.25W
C. 5.0W
D. 500W

Answer 1

Hint: By directly comparing the given value of V and I by $V = {V_o}\sin \omega t $ and $ I = {I_o}\sin (\omega t + \phi )$ we can find the values of \[{V_o}\], \[{I_o}\] and $\phi $ to apply the power dissipated formula i.e. $P = Vrms \times Irms \times \cos \phi $ to determine the correct option.

Formula used: $P = Vrms \times Irms \times \cos \phi $

Complete Step-by-Step solution:
The given values are V = 50sin50t and I =$100\sin (50t + \dfrac{\pi }{3})mA$.
Comparing the given equation of V by by general equation of $V = {V_o}\sin \omega t$
and we get \[{V_o} = 50volts\]
Now comparing the given equation of I from general equation of $I = {I_o}\sin (\omega t + \phi )$ and we get ${I_o}$= 100 also $\phi = \dfrac{\pi }{3}$
Dissipated power, $P = Vrms \times Irms \times \cos \phi $
Here, $Vrms = \dfrac{{{V_o}}}{{\sqrt 2 }} = \dfrac{{50}}{{\sqrt 2 }};Irms = \dfrac{{{I_o}}}{{\sqrt 2 }} = \dfrac{{100}}{{\sqrt 2 }}and\operatorname{Cos} \phi = $$Cos\dfrac{\pi }{3} = \dfrac{1}{2}$
Thus, $p = \dfrac{{50}}{{\sqrt 2 }} \times \dfrac{{100}}{{\sqrt 2 }} \times \dfrac{1}{2} = 1250mW = 1.25W$

Note: In AC circuit the instantaneous values of the voltage, current and therefore power are constantly changing being influenced by the supply AC circuits contain reactance, so there is a power component as a result of the magnetic and/or electric fields created by the components thus this result as unlike a purely resistive component, this power is stored and then returned back to the supply as the sinusoidal waveform goes through one complete periodic cycle.