Answer:
The question is incomplete.the complete complete question is giving below " A 60-Hz, single-phase source with V=277<30⁰ volts is applied to a circuit element. (a) Determine the instantaneous source voltage. Also determine the phasor and instantaneous currents entering the positive terminal if the circuit element is (b) a 20-Ω resistor, (c) a 10-mH inductor, and (d) a capacitor with 25-Ω reactance.
a. V=391.7cos(wt+30)V
b. I=19.6cos(wt+30)A
c. I=103.9cos(wt-60)A
d. I=15.7cos(wt+120)A
Explanation:
Recall the general expression for instantaneous voltage, giving as
V=Vₙcos(wt+Φ)
where vₙ is the amplitude. to calculate the amplitude, we use the equation below
Vₙ=√(2)*V
Hence the amplitude voltage is
Vₙ=√(2)*277
Vₙ=391.7v
and the angle Φ=30
Hence the instantaneous voltage is
V=391.7cos(wt+30)v
b. To determine the current entering the 20ohms resistor, we use the ohms law
I=Vₙ/R
I=391.7/20
I=19.6A
Note since the resistance is offered by only a resistor, it does not have effect on the phasor angle, hence the current is
I=19.6cos(wt+30)A
c. For the 10-mH inductor, we solve for the reactance as shown below
Xl=2Πfl
where f=60Hz
Xl=2π*60*10*10⁻³
Xl=3.78ohms
the current is thus
I=Vₙ/Xl
I=391.7/3.78
I=103.9A
the phasor angle will lag behinde by 90⁰, so
Φ=30-90
Φ=-60
I=103.9cos(wt-60)A
d. for a capacitor with reactance 25 ohms, the current is
I=Vₙ/Xc
I=391.7/25
I=15.7A
the phasor angle will lead by 90⁰, so
Φ=30+90
Φ=120
I=15.7cos(wt+120)A
