- #1
PhysicsTest
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- 26
- Homework Statement
- For the circuit element , calculate (i)the instantaneous power absorbed, (ii) the real power (state whether it is delivered or absorbed), (iii) the reactive power (state whether delivered or absorbed), (iv) the power factor (state whether lagging or leading)
- Relevant Equations
- p = v*i
The voltage and current are given as ##v(t) = 359.3\sin(\omega t + 15) volts ## and current is ## i(t) = 100\cos(\omega t + 5) ##. My question here is do i need to convert the voltage into cos terms or current into sin terms. suppose if i convert the voltage into cos terms, the calculations will look like
##v(t) = 359.3\cos(\omega t -75); ##
##i(t) = 100 \cos(\omega t + 5);##
The general formula for ##v(t) = V_{max} \cos(\omega t+\delta) ## and ##i(t) = I_{max} \cos(\omega t+\beta)##
##P(t) = VI_{R}[1 + \cos[2\omega t + \delta]] + VI_{X}\sin(2(\omega t + \delta))##
##V = \frac{V_{max}} {\sqrt 2} ##
##I_R = VI\cos(\delta - \beta) = 359.3 * 100/2 *\cos(-75-5) = 3119 ##
##I_X = VI \sin(\delta - \beta) = -17692 ##
##P(t) = 3119(1 + \cos(2\omega t -75))-17692(\sin(2\omega t -75))##
Is it correct?
##v(t) = 359.3\cos(\omega t -75); ##
##i(t) = 100 \cos(\omega t + 5);##
The general formula for ##v(t) = V_{max} \cos(\omega t+\delta) ## and ##i(t) = I_{max} \cos(\omega t+\beta)##
##P(t) = VI_{R}[1 + \cos[2\omega t + \delta]] + VI_{X}\sin(2(\omega t + \delta))##
##V = \frac{V_{max}} {\sqrt 2} ##
##I_R = VI\cos(\delta - \beta) = 359.3 * 100/2 *\cos(-75-5) = 3119 ##
##I_X = VI \sin(\delta - \beta) = -17692 ##
##P(t) = 3119(1 + \cos(2\omega t -75))-17692(\sin(2\omega t -75))##
Is it correct?