A 60-Hz, single-phase source with V 5 277/308 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.

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