Question

Consider the series RLC circuit of Problem 2.7 and calculate the complex power absorbed by each of the R, L, and C elements, as well as the complex power absorbed by the total load. Draw the resultant power triangle. Check whether the complex power delivered by the source equals the total complex power absorbed by the load.

Short answer

(a)

Consider the circuit diagram according to the given information.

(b)

The value of series impedance of the circuit is$5\angle 53.13°\mathrm{\Omega }$.

(c)

The value of current I delivered by the voltage source is$20\angle -53.13°\text{\hspace{0.33em}A}$.

The value of complex power absorbed by the resistor $S_{\text{R}}$is $\text{1200\hspace{0.33em}VA}$.

The value of complex power absorbed by the inductor $S_L$is $3200\angle 90°\text{\hspace{0.33em}VA}$.

The value of complex power absorbed by the element $S_C$is $1600\angle -90°\text{\hspace{0.33em}VA}$.

The value of complex power absorbed by the total load $S_{load}$is $1200+j1600\text{\hspace{0.33em}VA}$.

The value of complex power delivered by the source $S_{\text{source}}$is $1200+j1600\text{\hspace{0.33em}VA}$.

Consider the circuit diagram of resultant power triangle.

Step-by-step solution

Write the given data from the question.

Consider the reference of problem 2.7.

The value of given source voltage is $V=100\angle 0°\text{\hspace{0.33em}V}$.

The Value of given resistor is $\text{R}=3\mathrm{\Omega }$.

The value of given inductive reactance is $X_{\text{L}}=j8\mathrm{\Omega }$.

The value of given capacitive reactance is $X_{\text{c}}=-j4\mathrm{\Omega }$.

Determine the formula of series impedance, current and complex power.

Write the formulaof series impedance of the circuit.

${\mathbf{Z}}_S=\mathbf{R}\mathbf{+}{\mathbf{X}}_L\mathbf{+}{\mathbf{X}}_C$ …… (1)

Here, $\mathbf{R}$is resistance, ${\mathbf{X}}_L$is inductive reactance and ${\mathbf{X}}_C$is capacitive reactance.

Write the formula of current delivered by the voltage source.

$\mathbf{I}=\frac{V}{{\mathbf{Z}}_S}$ …… (2)

Here, $\mathbf{V}$is voltage and ${\mathbf{Z}}_S$is source impedance.

Write the formula of complex power absorbed by the resistor ${\mathrm{S}}_{\mathrm{R}}$.

${\mathbf{S}}_R=\left(\mathrm{IR}\right){\mathbf{I}}^{*}$ …… (3)

Here, I is current and R is resistance.

Write the formula of complex power absorbed by the inductor $S_L$.

${\mathbf{S}}_L=\left({\mathbf{IX}}_L\right){\mathbf{I}}^{*}$ …… (4)

Here, I is current and ${\mathbf{X}}_{\mathbf{L}}$is inductive reactance.

Write the formula of complex power absorbed by the element $S_c$.

${\mathbf{S}}_C=\left({\mathbf{IX}}_C\right){\mathbf{I}}^{*}$ …… (5)

Here, I is current and ${\mathbf{X}}_C$is capacitive reactance.

Write the formula of complex power absorbed by the total load .

${\mathbf{S}}_{\mathrm{load}}={\mathbf{S}}_R\mathbf{+}{\mathbf{S}}_L\mathbf{+}{\mathbf{S}}_C$ …… (6) Here, ${\mathbf{S}}_R$is complex power absorbed by the resistor, ${\mathbf{S}}_L$is complex power absorbed by the inductor and ${\mathbf{S}}_C$is complex power absorbed by the element.

Write the formula of complex power delivered by the source $S_{\text{source}}$.

${\mathbf{S}}_{\mathrm{source}}={\mathbf{VI}}^{*}$ …… (7)

Here, V is voltage and ${\mathbf{I}}^{*}$is current source.

(a) Draw a circuit diagram according to the given information.

Draw a circuit diagram to the given information.

(b) Determine the series impedance ZS of the circuit.

Determine the expression of series impedance $Z_S$.

Substitute $\mathit{j}\mathbf{8}$for ${\mathbf{X}}_L$, 3 for R and $-j4$for $X_{\text{C}}$into equation (1).

$\begin{array}{c}{Z}_S=3+j8-j4\\ =3+j4\\ =5\angle 53.13°\mathrm{\Omega }\end{array}$

Therefore, the characteristics impedance $Z_S$is $5\angle 53.13°\Omega$.

(c) Determine the current I delivered by the voltage source.

Determine the value of current I delivered by the voltage source.

Substitute $100\angle 0°$for V and $5\angle 53.13°\text{\hspace{0.33em}A}$for $Z_S$into equation (2).

$\begin{array}{c}I=\frac{100\angle 0°}{5\angle 53.13°}\\ =20\angle -53.13°\text{\hspace{0.33em}A}\end{array}$

Determine thecomplex power absorbed by the resistor R.

Substitute $20\angle -53.13°$ for I, 3 for R and $20\angle -53.13°$for I into equation (3).

$\begin{array}{c}{S}_{\text{R}}=\left(20\angle -53.13°\right)\left(3\right){\left(20\angle -53.13°\right)}^{*}\\ =\left(60\angle -53.13°\right)\left(20\angle 53.13°\right)\\ =1200\text{\hspace{0.33em}VA}\end{array}$

Therefore, the complex power absorbed by the resistor $S_{\text{R}}$is $1200\mathrm{VA}$.

Determine the complex power absorbed by the inductor L.

Substitute $20\angle -53.13°$ for I, j8 for $X_L$and $20\angle -53.13°$for $I^{*}$into equation (4).

$\begin{array}{c}{S}_{\text{L}}=\left(20\angle -53.13°\right)\left(j8\right){\left(20\angle -53.13°\right)}^{*}\\ =\left(20\angle -53.13°\right)\left(8\angle 90°\right)\left(20\angle 53.13°\right)\\ =3200\angle 90°\text{\hspace{0.33em}VA}\end{array}$

Therefore, the complex power absorbed by the inductor $S_{\text{L}}$is $3200\angle 90°\text{\hspace{0.33em}VA}$.

Determine the complex power absorbed by the element C.

Substitute $20\angle -53.13°$ for I, $-j4$for $X_{\text{C}}$and $20\angle -53.13°$for $I^{*}$into equation (5).

$\begin{array}{c}{S}_{\text{C}}=\left(20\angle -53.13°\right)\left(-j4\right){\left(20\angle -53.13°\right)}^{*}\\ =\left(20\angle -53.13°\right)\left(4\angle -90°\right)\left(20\angle 53.13°\right)\\ =1600\angle -90°\text{\hspace{0.33em}VA}\end{array}$

Therefore, the complex power absorbed by the element $S_c$is $1600\angle -90°\text{\hspace{0.33em}VA}$.

Determine thecomplex power absorbed by the total load.

Substitute 1200 for $S_R$, $3200\angle 90°$for $S_L$and $1600\angle -90°$for $S_c$into equation (6).

$\begin{array}{c}{S}_{\text{load}}=1200+3200\angle 90°+1600\angle -90°\\ =2000\angle 53.13°\\ =1200+j1600\text{\hspace{0.33em}VA}\end{array}$

Therefore, the complex power absorbed by the total load $S_{\text{load}}$is $1200+j1600\text{\hspace{0.33em}VA}$.

Determine thecomplex power delivered by the source.

$\begin{array}{c}{S}_{\text{source}}=\left(100\angle 0°\right){\left(20\angle -53.13°\right)}^{*}\\ =\left(100\angle 0°\right)\left(20\angle 53.13°\right)\\ =2000\angle 53.13°\\ =1200+j1600\text{\hspace{0.33em}VA}\end{array}$

Therefore, complex power delivered by the source $S_{source}$is $1200+j1600\text{\hspace{0.33em}VA}$.

Hence, from above result obtained the conclusion that the complex power delivered by source is same as complex power absorbed by load.

Therefore, the source complex power is $S_{\text{source}}=S_{\text{load}}$.

Draw the circuit diagram of resultant power triangle.