Chapter 11, Problem 61P

Given the circuit in Fig. 11.80, find I_o and the overall complex power supplied.

To determine

Calculate the current $I_o$ and the overall complex power supplied of the circuit shown in Figure 11.80.

Answer to Problem 61P

The current $I_o$ is $66.2\angle 92.4°\text{A}$ and the overall complex power supplied is $S_o$ is $6.62\angle -2.4°\text{kVA}$.

Explanation of Solution

Given data:

Refer to Figure 11.80 in the textbook.

The current $V_o$ is $100\angle 90°\text{V}$.

For load A,

The apparent power $S$ is $2\text{kVA}$.

The power factor $\text{pf}$ is $0.707\left(\text{leading}\right)$.

For load B,

The real power $P$ is $1.2\text{kW}$.

The real power $Q$ is $0.8\text{kVAR}\text{(capacitive)}$.

For load C,

The real power $P$ is $4\text{kW}$.

The power factor $\text{pf}$ is $0.9\left(\text{lagging}\right)$.

Formula used:

Write the expression to find the complex power.

$S=P+jQ$ (1)

Here,

$P$ is the real power, and

$Q$ is the reactive power.

Write the expression to find the power factor $\text{pf}$.

$\text{pf}=\cos \left(\theta \right)$ (2)

Here,

$\theta$ is the phase angle.

Write the expression to find the real power.

$P=S\cos \left(\theta \right)$ (3)

Write the expression to find the reactive power.

$Q=S\sin \theta$ (4)

Write the expression to find the output voltage.

$I^{*}=\frac{S}{V}$ (5)

Calculation:

The given Figure 11.80 is redrawn as shown in Figure 1.

For load A:

Substitute $0.707$ for $\text{pf}$ in equation (2) to find the phase angle.

$\begin{array}{l}0.707=\cos \left(\theta \right)\\ \theta ={\cos }^{-1}\left(0.707\right)\\ \theta =45°\end{array}$

Substitute $2\text{kVA}$ for S and $36.87°$ for $\theta$ in equation (3) to find the real power P in watts.

$\begin{array}{c}P=2\text{kVA}\cos \left(45°\right)\\ =1.4142\text{kW}\left\{\because 1\text{W}=\text{V}\cdot \text{A}\right\}\end{array}$

Substitute $2\text{kVA}$ for S and $45°$ for $\theta$ in equation (4) to find the reactive power Q in VAR.

$\begin{array}{c}Q=2\text{kVA}\sin \left(45°\right)\\ =1.4142\text{kVAR}\end{array}$

Substitute $1.4142\text{kW}$ for P and $1.4142\text{kVAR}$ for $Q$ in equation (1) to find the complex power in VA.

$S_A=\left(1.4142+j1.4142\right)\text{kVA}$

As the power factor is leading, the load is capacitive. Therefore, the equation becomes,

$S_A=\left(1.4142-j1.4142\right)\text{kVA}$

For load B:

Substitute $1.2\text{kW}$ for P and $0.8\text{kVAR}$ for $Q$ in equation (1) to find the complex power in VA.

$S_B=\left(1.2+j0.8\right)\text{kVA}$

As the load is capacitive, the power factor is leading. Therefore, the equation becomes,

$S_B=\left(1.2-j0.8\right)\text{kVA}$

For load C:

Substitute $0.9$ for $\text{pf}$ in equation (2) to find the phase angle.

$\begin{array}{l}0.9=\cos \left(\theta \right)\\ \theta ={\cos }^{-1}\left(0.9\right)\\ \theta =25.84°\end{array}$

Substitute $25.84°$ for $\theta$ and $4\text{kW}$ for P in equation (3) to find the apparent power S in VA.

$4\text{kW}=S\cos \left(36.87°\right)$

Rearrange the equation as follows,

$\begin{array}{c}S=\frac{4\text{kVA}}{\cos \left(25.84°\right)}\left\{\because 1\text{W}=1\text{V}\cdot 1\text{A}\right\}\\ =4.444\text{kVA}\end{array}$

Substitute $4.444\text{kVA}$ for S and $25.84°$ for $\theta$ in equation (4) to find the reactive power Q in VAR.

$\begin{array}{c}Q=4.444\text{kVA}\sin \left(25.84°\right)\\ =1.937\text{kVAR}\end{array}$

Substitute $4\text{kW}$ for P and $1.937\text{kVAR}$ for $Q$ in equation (1) to find the complex power in VA.

$S_C=\left(4+j1.937\right)\text{kVA}$

In Figure 1, the load B and load C are connected in series. Therefore,

$S_D=S_B+S_C$

Substitute $\left(1.2-j0.8\right)\text{kVA}$ for $S_B$ and $\left(4+j1.937\right)\text{kVA}$ for $S_C$ in the equation to find the complex power in VA.

$\begin{array}{c}{S}_D=\left(1.2-j0.8\right)\text{kVA}+\left(4+j1.937\right)\text{kVA}\\ =\left(5.2+j1.137\right)\text{kVA}\end{array}$

The modified Figure is shown in Figure 2.

Substitute $\left(5.2+j1.137\right)\text{kVA}$ for $S_D$ and $100\angle 90°\text{V}$ for $V_o$ in equation (5) to find the current $I_2$ in amperes.

$\begin{array}{l}{I}_2^{*}=\frac{\left(5.2+j1.137\right)\times {\text{10}}^3\text{VA}}{100\angle 90°\text{V}}\left\{\because 1\text{k}={10}^3\right\}\\ I_2^{*}=\left(11.37-j52\right)\text{A}\\ I_2=\left(11.37+j52\right)\text{A}\end{array}$

Substitute $\left(1.4142-j1.4142\right)\text{kVA}$ for $S_A$ and $100\angle 90°\text{V}$ for $V_o$ in equation (5) to find the current $I_1$ in amperes.

$\begin{array}{l}{I}_1^{*}=\frac{\left(1.4142-j1.4142\right)\times {\text{10}}^3\text{VA}}{100\angle 90°\text{V}}\left\{\because 1\text{k}={10}^3\right\}\\ I_1^{*}=\left(-14.142-j14.142\right)\text{A}\\ I_1=\left(-14.142+j14.142\right)\text{A}\end{array}$

Apply Kirchhoff’s current law in Figure 2 to find the current $I_o$.

$I_o=I_1+I_2$

Substitute $\left(-14.142+j14.142\right)\text{A}$ for $I_1$ and $\left(11.37+j52\right)\text{A}$ for $I_2$ in the equation to find the current $I_o$ in amperes.

$\begin{array}{c}{I}_o=\left(-14.142+j14.142\right)\text{A}+\left(11.37+j52\right)\text{A}\\ =\left(-2.772+j66.14\right)\text{A}\end{array}$

Convert the equation from rectangular to polar form.

$I_o=66.2\angle 92.4°\text{A}$

The overall complex power supplied by the source is,

$S_o=V_o{I}_o^{*}$

Substitute $100\angle 90°\text{V}$ for $V_o$ and $66.2\angle 92.4°\text{A}$ for $I_o$ to find the overall complex power in VA.

$\begin{array}{c}{S}_o=\left(100\angle 90°\text{V}\right){\left(66.2\angle 92.4°\text{A}\right)}^{*}\\ =\left(100\angle 90°\right)\left(66.2\angle -92.4°\right)\text{VA}\\ =\text{6620}\times {\text{10}}^3\times {\text{10}}^{-3}\angle -\text{2}\text{.4}\text{VA}\\ =6.62\angle -\text{2}\text{.4}°\text{kVA}\left\{\because 1\text{k}={10}^3\right\}\end{array}$

Conclusion:

Thus, the current $I_o$ is $66.2\angle 92.4°\text{A}$ and the overall complex power supplied is $S_o$ is $6.62\angle -2.4°\text{kVA}$.