Question

(a) Transform$\mathbf{V}\mathbf{}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{75}\mathbf{cos}\mathbf{(}\mathbf{377}\mathbf{t}\mathbf{-}{\mathbf{15}}^{\mathbf{0}\mathbf{}}\mathbf{)}$to phasor form. Comment on whether $\mathbf{\omega }\mathbf{=}\mathbf{377}$appears in your answer. (b) Transform $\mathbf{V}\mathbf{=}\mathbf{50}\mathbf{\angle }{\mathbf{10}}^{\mathbf{0}}$to instantaneous form. Assume that $\mathbf{\omega }\mathbf{=}\mathbf{377}$(c) Add the two sinusoidal functions a(t) and b(t)andof the same frequency given as follows: a(t) = A$\sqrt{\mathbf{2}}$$\mathbf{cos}\mathbf{(}\mathbf{\omega t}\mathbf{+}\mathbf{a}\mathbf{)}$and b(t) = B$\sqrt{\mathbf{2}}$$\mathbf{cos}\mathbf{(}\mathbf{\omega t}\mathbf{+}\mathbf{\beta }\mathbf{)}$. Use phasor methods and obtain the resultant c(t) . Does the resultant have the same frequency?

Short answer

(a) No, the angular frequency $\mathrm{\omega }=377$is not appearing in the phasor form of voltage.

(b) The instantaneous form of voltage is $\mathrm{V}\left(\mathrm{t}\right)=70.71\cos \left(377\mathrm{t}+{10}^0\right)$.

(c)The resultant value of c(t) is $\sqrt{2}{\mathrm{C}}_{\mathrm{ms}}\cos \left(\mathrm{\omega t}+\mathrm{\delta }\right)$and have same frequency of $\mathrm{\omega }$.

Step-by-step solution

Write the given data of the question

The instantaneous value of voltage is, v(t) = 75 cos(377t - 15^o) .

The value of the angular frequency is, $\mathrm{\omega }=377$rad/sec .

The phasor form of voltage is $50\angle {10}^0\mathrm{V}$.

Write the concept for polar form;

If x(t) = Acost is taken as the reference for the phasor diagram.

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

$\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{A}}{\sqrt{\mathbf{2}}}\mathbf{\angle }\mathbf{-}\mathbf{\theta }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

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

The expression for the instantaneous voltage is given by,

$\mathbf{v}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{v}}_{\mathbf{max}}\mathbf{cos}\mathbf{(}\mathbf{\omega t}\mathbf{+}\mathbf{\delta }\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Here v(t) is the instantaneous voltage,${\mathbf{V}}_{\mathbf{max}}$is the maximum voltage, $\mathbf{\omega }$is the angular frequency and $\mathbf{\delta }$is the phase angle.

The maximum value of the voltage is given by,

${\mathbf{V}}_{\mathbf{max}}\mathbf{=}\sqrt{\mathbf{2}}{\mathbf{V}}_{\mathbf{rms}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Here V_rmsis the rms voltage and V_maxis the peak value of the voltage.

Calculate phasor form of the given supply voltage.

(a)

The rms value of the voltage is given by,

${\mathrm{V}}_{\mathrm{rms}}=\frac{{\mathrm{V}}_{\max }}{\sqrt{2}}$

Here V_rmsis the rms voltage and V_max is the peak value of the voltage.

Substitute 75V for V_maxin the equation.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{rms}}=\frac{75}{\sqrt{2}} \\ =53.03\mathrm{V} \end{gathered}$$

Then the phasor form of voltage is $53.03\angle -{15}^0\mathrm{V}$.

Therefore the angular frequency $\mathrm{\omega }$= 377rad/sec is not appearing in the phasor form of voltage.

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

(b)

The phasor form of voltage is given by,

$\mathrm{V}=50\angle {10}^0\mathrm{V}$.

The rms value of the voltage is given by,

V_rms = 50 V

The maximum value of the voltage is given by,

${\mathrm{V}}_{\max }=\sqrt{2}{\mathrm{V}}_{\mathrm{rms}}$

Here V_rms is the rms voltage and V_maxis the peak value of the voltage.

Substitute 50 V for V_rms in the equation.

$$\begin{gathered} {\mathrm{V}}_{\max }=\sqrt{2}\left(50\right) \\ =70.71\mathrm{V} \end{gathered}$$

The expression for the instantaneous voltage is given by,

$\mathrm{V}\left(\mathrm{t}\right)={\mathrm{V}}_{\max }\cos \left(\mathrm{\varpi t}+\mathrm{\delta }\right)$

Here v (t) is the instantaneous voltage, V_maxis the maximum voltage, $\mathrm{\omega }$is the angular frequency and $\delta$is the phase angle.

Substitute 70.71 V for V_max, 377 rad/sec for $\mathrm{\omega }$ and 10^ofor $\mathrm{\delta }$in the equation.

$\mathrm{V}\left(\mathrm{t}\right)=70.71\cos \left(377t+{10}^0\right)$

Therefore the instantaneous form of voltage is v (t) = 70.71 cos (377t + 10^o ) .

Calculate the resultant c(t)  .

(c)

The phasor form c (t) can be written as,

$$\begin{gathered} \mathrm{c}\left(\mathrm{t}\right)=\mathrm{a}\left(\mathrm{t}\right)+\mathrm{b}\left(\mathrm{t}\right) \\ =\mathrm{A}\sqrt{2}\cos \left(\mathrm{\varpi t}+\mathrm{a}\right)+\mathrm{B}\sqrt{2}\cos \left(\mathrm{\varpi t}+\mathrm{\beta }\right) \end{gathered}$$

Above expression can be written in phasor form as,

$C=\mathrm{A}\angle \mathrm{a}+\mathrm{B}\angle \mathrm{\beta }$

Expand the expression in sin and cos form.

$$\begin{gathered} C=\mathrm{A}\angle \mathrm{\alpha }+\mathrm{B}\angle \mathrm{\beta } \\ =(\mathrm{Acos}\mathrm{\alpha }+\mathrm{jAsin}\mathrm{\alpha })+(\mathrm{Bcos}\mathrm{\beta }+\mathrm{jBsin}\mathrm{\beta }) \\ =(\mathrm{Acos}\mathrm{\alpha }+\mathrm{Bcos\beta })+\mathrm{j}(\mathrm{Asin}\mathrm{\alpha }+\mathrm{Bsin}\mathrm{\beta }) \\ =C_{\mathrm{rms}}\angle \mathrm{\delta } \end{gathered}$$

The maximum value of the c is given by,

${\mathrm{C}}_{\max }=\sqrt{2}{\mathrm{c}}_{\mathrm{rms}}$

Here C_msis the rms value and C_maxis the peak value.

The expression for the instantaneous of C is given by,

$\mathrm{C}\left(\mathrm{t}\right)={\mathrm{C}}_{\max }\cos \left(\mathrm{\varpi t}+\mathrm{\delta }\right)$

Here c(t) is the instantaneous value, C_max is the maximum value, $\mathrm{\omega }$is the angular frequency and $\mathrm{\delta }$is the phase angle.

Substitute $\sqrt{2}$C_msfor C_maxn the equation.

$\mathrm{C}\left(\mathrm{t}\right)=\sqrt{2}{\mathrm{C}}_{\mathrm{rms}}\cos \left(\mathrm{\varpi t}+\mathrm{\delta }\right)$

Therefore the resultant value of c (t) is $\sqrt{2}$C_mscos( $\mathrm{\omega }$t +$\mathrm{\delta }$).