Question

A60Hz single –phase with V=277$\angle$30^ovolts is applied to a circuit element. (a) Determine the instantaneous source voltage. Also determine the phasor and instantaneous currents entering the positive terminal if the circuit element is (b) 20$\mathrm{\Omega }$resistor (c) a 10-mH inductor, and (d) a capacitor with 25-$\mathrm{\Omega }$ reactance.

Short answer

(a)The instantaneous source voltage is 391.74 cos (377t + 30^o) V.

(b)The phasor current entering the positive terminal of the resistor is 13.85$\angle$30^oA and the instantaneous current is 19.587 cos (377t + 30^o)A

(c)The phasor current entering the positive terminal of the resistor is 73.476 $\angle$-60^oA and the instantaneous current is 103.91 cos(377t - 60^o)A

(d)The phasor current entering the positive terminal of the resistor is 11.08$\angle$120^oA and the instantaneous current is 15.67 cos (377t + 120^o)A

Step-by-step solution

Write the given data of the question

The source voltage is given in phasor form V = 277$\angle$30^o.

The value of the supply frequency is f = 60Hz.

The value of the Resistor is R = 20$\mathrm{\Omega }$.

The value of the Inductor is L= 10 mH.

The Capacitive reactance is X _c= 25$\mathrm{\Omega }$.

Write the concept for polar form;

If x (t) = Acos$\mathrm{\omega }$t is taken as the reference for the phasor diagram.

The polar form or phasor expression of y(t) =Acos($\mathrm{\omega }$t -$\mathrm{\theta }$) is,

Here, -$\mathrm{\theta }$means that y(t) lags x(t) by $\mathrm{\theta }$angle, A is the maximum or peak value of y (t) and $\frac{\mathrm{A}}{\sqrt{2}}$is the rms value of the y (t) .

Calculate the instantaneous voltage.

(a)

Consider the phasor voltage is,

V = 277 ∠ 30^o

The amplitude or peak value of the source voltage is given by,

V_{m = √2}V_rms

Here, V_mis the peak value of the supply voltage, and V_rms rms (root mean square) value of the supply voltage.

Substitute 277V for V_rms in the equation.

Write the expression for the instantaneous voltage.

Here, V_mis the peak value of the source voltage, f is the supply frequency, t is time duration, v (t) is the instantaneous voltage and $\mathrm{\theta }$is the angle through which supply voltage may lag or lead from the reference value.

Substitute 391.74 V for V_m , 60Hz for f and 30^o for $\mathrm{\theta }$in above equation.

Therefore, the instantaneous value for the source voltage is 391.74cos (377t + 30^o ).

Calculate phasor current and instantaneous current entering the positive terminal of the resistor.

(b)

Write the phasor current entering the positive terminal.

Here, /_R is the current entering the resistor, V is the voltage across the resistor and R is the resistance.

Substitute 277$\angle$30^ofor V and 20 - $\mathrm{\Omega }$for R in the equation.

The instantaneous current is given by

Substitute 13.85 √2 A for /_RM , 60Hz for f and 30^ofor θin the above equation.

Therefore, the phasor current entering the resistor is 13.85$\angle$30^oand the instantaneous current is 19.587 cos (377t + 30^o).

Calculate phasor current and instantaneous current entering the positive terminal of the inductor.

(c)

The inductive reactance value is given by,

X_L= j2$\mathrm{\pi }$fL

Here, X_L is the inductive reactance, f is the frequency and L is the inductance.

Substitute 60Hz for f and 10 localid="1656766275678" $\times$10^-3H for L in the above equation.

The phasor current entering the positive terminal of the resistor is;

Here, /_Lis the current entering the inductor, V is the voltage across the inductor and X_Lis the inductive reactance.

Substitute 277localid="1656766281358" $\angle$30^ofor V and j3.77 localid="1656766286738" $\mathrm{\Omega }$forX_L in above equation.

The instantaneous current is given by,

Substitute 73.476localid="1656766292116" $\sqrt{2}$A for /_LM, 60Hz for f and -60^oforlocalid="1656766298604" $\mathrm{\theta }$in above equation.

Therefore, the phasor current entering the inductor is 73.476$\angle$-60^oA and the instantaneous current is 103.91cos (377 t - 60^o) A.

Calculate phasor current and instantaneous current entering the positive terminal of the capacitor.

(d)

The capacitive reactance is given by,

The phasor current entering the positive terminal of the Capacitor is,

Here, /_cis the current entering the capacitor, V is the voltage across the inductor and X_cis the capacitive reactance.

Substitute 277$\angle$ 30^ofor V and -j25$\mathrm{\Omega }$for X_c .

Write the expression for the instantaneous current.

Substitute 11.08$\sqrt{2}$A for /_LM, 60Hz for f and 120^ofor$\mathrm{\theta }$in the above equation.

Therefore, the phasor current entering the inductor is 11.08$\angle$120^oA and the instantaneous current is 15.67cos (377 t + 120^o)A.