Question

Let a voltage source v (t) = 4cos (ωt +60^o) be connected to an impedance Z =3∠-45^oΩ .

(a) Given the operating frequency to be $\mathbf{60}\mathbf{Hz}$, determine the expressions for the current and instantaneous power delivered by the source as functions of time.

(b) Plot these functions along with $\mathbf{v}\left(t\right)$on a single graph for comparison.

(c) Find the frequency and average value of the instantaneous power.

Short answer

(a) The instantaneous current is 1.33cos (ωt +105^o) A and instantaneous power is 2.665 [cos(2ωt +165^o)+0.707] W

(b) The graph is drawn below:

(c) The frequency is 120Hz and average power is 1.885 W .

Step-by-step solution

Given data.

Thevalue of voltage source is,

v (t) = 4cos (ωt +60^o) V .........(1)

The impedance value is$Z=3\angle -45o\Omega$

The operating frequency is$\mathrm{f}=60\mathrm{HZ}$

Determine the formulas ofinstantaneous current, instantaneous power, and average power.

Write the formula of instantaneous current.

$\mathbf{i}\left(t\right)\mathbf{=}\frac{\mathbf{v}\left(t\right)}{\mathbf{z}}$.............(1)

Here, v (t) is ainstantaneous voltage and Z is impedance.

Write the formula of instantaneous power.

role="math" localid="1655114017652" $\mathbf{p}\left(t\right)\mathbf{=}\mathbf{v}\left(t\right)\mathbf{}\mathbf{i}\mathbf{}\left(t\right)$ ..............(2)

Here, v (t) is a instantaneous voltage and i (t)

is instantaneous current.

Write the formula of average power.

${\mathbf{p}}_{\mathbf{avg}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{}{\mathbf{V}}_{\mathbf{m}}\mathbf{}{\mathbf{I}}_{\mathbf{m}}\mathbf{cos}\mathbf{(}\mathbf{\theta }\mathbf{-}\mathbf{f}\mathbf{)}$

Here, V_m is a peak voltage, l_m is a peak current, $\mathrm{\Phi }$is a phase shift of current

and θ is a phase shift of voltage.

Determine the instantaneous current and instantaneous power.

(a)

Substitute $4\cos \left(\omega t+60°\right)\mathrm{V}$for$v\left(t\right)$and $3\angle -45°\Omega$for $Z$in equation (1).

$$\begin{gathered} i\left(t\right)=\frac{4\cos \left(\omega t+60°\right)V}{3\angle -45°\Omega } \\ =1.33\cos \left(\omega t+60°+45°\right) \\ =1.33\cos \left(\omega t+105°\right)A \end{gathered}$$ …… (2)

Substitute $1.33\cos \left(\omega t+105°\right)A\mathrm{for}i\left(t\right)$and $4\cos \left(\omega t+60°\right)V$for$v\left(t\right)$in equation (2).

$$\begin{gathered} p\left(t\right)=\left[4\cos \left(\mathrm{\omega t}+60°\right)\mathrm{V}\right]\left[1.33\cos \left(\mathrm{\omega t}+105°\right)\mathrm{A}\right] \\ =5.33\cos \left(\mathrm{\omega t}+60°\right)\cos \left(\mathrm{\omega t}+105°\right)\mathrm{W} \\ =5.33\times \frac{1}{2}\left[\cos \left(\mathrm{\omega t}+60°+\mathrm{\omega t}+105°\right)+\cos \left(\mathrm{\omega t}+60°+\mathrm{\omega t}+105°\right)\right] \\ =2.665\left[\cos \left(2\mathrm{\omega t}+165°\right)+0707\right]\mathrm{W} \end{gathered}$$

Therefore, the instantaneous current is$1.33\cos \left(\mathrm{\omega t}+105°\right)\mathrm{A}$and instantaneous power is$2.665\left[\cos \left(2\mathrm{\omega t}+165°\right)+0.707\right]\mathrm{W}$

plot the graph for instantaneous current, voltage and power.

(b)

From equation (1), the peak value of voltage is 4V and phase shift is -60^o .

From equation (2), the peak value of current is1.33 Aand phase shift is -105^o .

From equation (3), the peak value of power is 2.665 W .

The frequency of current and voltage wave will be f=60Hz .

The time period will be,

$$\begin{gathered} t=\frac{1}{f} \\ =\frac{1}{60\mathrm{Hz}} \\ =0.01667\sec \end{gathered}$$

The frequency of power signal is double of 60Hz .

Plot instantaneous voltage v (t) , current i (t) and power p (t) in a single graph.

Determine the average power and frequency of power.

(c)

Determine the base impedance.

Substitute 1.33 A for l_m , 4 V for V_m60^o for θ and 105^o for $\Phi$ in equation (3).

$$\begin{gathered} p\left(t\right)=\frac{1}{2}\left(1.33\right)\left(4\right)\cos \left(60°-105°\right) \\ =1.885\mathrm{W} \end{gathered}$$

From instantaneous power, it is observed that the frequency of power is always double the frequency of current and voltage.

Therefore, frequency of power is 120Hz .

Therefore, the frequency is 120Hz and average power is 1.885 W.