Block 06 - Complex Power

Learning objectives

Preparation at Home

Well, again

• read through the present chapter and write down anything you did not understand.
• Also here, there are some clips for more clarification under 'Embedded resources' (check the text above/below, sometimes only part of the clip is interesting).

For checking your understanding please do the following exercises:

• …

90-minute plan

1. Warm-up (x min):
   1. ….
2. Core concepts & derivations (x min):
   1. …
3. Practice (x min): …
4. Wrap-up (x min): Summary box; common pitfalls checklist.

Conceptual overview

Core content

Last semester in block18 we investigated induciton effects have into coils. When the coil is rotating constantly, These can be considered as the AC voltace source on a circuit:

• the sinusoidal alternating voltage is produced by the rotation of a coil in a homogeneous magnetic field, and
• the sinusoidal alternating current is formed by a connected load (or complex impedance).

This will be briefly illustrated here. In figure 1 a coil with $w$ windings is seen in a magnetic field with a magnetic flux density $\vec{B}$. The coil rotates - starting from $\varphi_0$ with angular velocity $\omega$. The rotation changes the chained flux $\Psi$ through the coil and thus a voltage $u(t)$ is induced.

For the rotation angle $\varphi$ holds: \begin{align*} \varphi(t)=\omega t + \varphi_0 \quad \text{with} \quad \varphi_0=\varphi(t=0) \\ \end{align*}

Thus, the induced voltage $u(t)$ is given by: \begin{align*} u(t) &= -\frac{{\rm d~} \Psi} {{\rm d}t} \\ &= -N \cdot \frac{{\rm d~} \Phi} {{\rm d}t} \\ &= -NBA\cdot \frac{{\rm d~} \cos \varphi(t)} {{\rm d}t} \\ &= -\hat{\Psi}\cdot\frac{{\rm {\rm d~}} \cos (\omega t + \varphi_0)}{{\rm d}t} \\ &= \omega \hat{\Psi} \cdot \sin (\omega t + \varphi_0) \\ &= \hat{U} \cdot \sin (\omega t + \varphi_0) \\ \end{align*}

Such single-phase systems are therefore alternating current systems, which use one outgoing line and one return line each for the current conduction.
Out of the last formula we derived the following instantaneous voltage $u(t)$ \begin{align*} u(t) &= \hat{U} \cdot \sin (\omega t + \varphi_0) \\ &= \sqrt{2} U\cdot \sin (\omega t + \varphi_0) \\ \end{align*}

In block03, we used DC power $P = U \cdot I$ and compared it with the instantaneous power $p(t)$ of an AC circuit:

\begin{align*} p(t) &= \color{blue}{u(t)} \cdot \color{red}{i(t)} \end{align*}

Additionally we used a bit later in block03 the functions representing the instantaneous signals to derive the inductance: $x(t)= \sqrt{2}{X}\cdot \sin(\omega t + \varphi_x)$

Now, we will combine both to analyze the AC power on the resistor, capacitor and inductivity in more detail.

Ideal Ohmic resistance R

The simplest component to look at for the instantaneous power is the resistor. For this, we start with the basic definition of the instantaneous voltage $u_R(t)$ (which was given in the last semester) as

\begin{align*} \color{blue}{u_R(t)} &= \sqrt{2}U \sin(\omega t + \varphi_u) \end{align*}

With the defining formula for the resistor, we get:

\begin{align*} \color{blue}{u_R(t)} &= R \cdot \color{red}{i(t)} \\ \rightarrow \color{red}{i(t)} &= {{\color{blue}{u_R(t)}}\over{R}} \\ &= \sqrt{2}{ {U} \over{R}} \sin(\omega t + \varphi_u) \end{align*}

This leads to an instantaneous power $p_R(t)$ of \begin{align*} p_R(t) &= \color{blue}{u_R(t)} \cdot \color{red}{i_R(t)} \\ &= 2\cdot {{U^2}\over{R}} \sin^2(\omega t + \varphi_u) \\ &= {{U^2}\over{R}}\left(1- \cos\left(2\cdot (\omega t + \varphi_u)\right)\right) \\ \end{align*}

For the last step the Double-angle formula “$\cos(2x) = 1 - 2 \sin^2(x)$” was used.

This result is interesting in the following ways:

1. The part $1- \cos (2\cdot (\omega t + \varphi_u) )$ is always non-negative and a shifted sinusoidal function between $0...2$. The average value of this part is $1$.
2. The average value of $p_R(t)$ is then: $P_R = {{U^2}\over{R}}$
3. The use of the $\sqrt{2}$ in the definition $\color{blue}{u_R(t)} = \sqrt{2} U \sin(\omega t + \varphi_u)$ leads to the average power as $P_R = {{U^2}\over{R}}$. This formula for the power is exactly like the formula for the power in pure DC situations.

Ideal Capacity C

Also here, we start with the basic definition of the instantaneous voltage

\begin{align*} \color{blue}{u_C(t)} &= \sqrt{2}U \sin(\omega t + \varphi_u) \end{align*}

With the defining formula for the capacity, we get: \begin{align*} \color{red}{i_C(t)} &= C {{{\rm d}\color{blue}{u_C(t)}}\over{{\rm d}t}} \\ &= \sqrt{2} U \omega C \cos(\omega t + \varphi_u) \end{align*}

This leads to an instantaneous power $p_C(t)$ of

\begin{align*} p_C(t) &= \color{blue}{u_C(t)} \cdot \color{red}{i_C(t)} \\ &= 2\cdot U^2 \omega C \cdot \sin( \omega t + \varphi_u) \cos(\omega t + \varphi_u) \\ &= + U^2 \omega C \cdot \sin( 2\cdot (\omega t + \varphi_u)) \\ \end{align*}

Also, this result is interesting:

1. The part $\sin( 2\cdot (\omega t + \varphi_u))$ has an average value of $0$.
2. Therefore, also the average value of $p_C(t)=0$

Ideal Inductivity L

A similar approach is done for the ideal inductivity. We again start with the basic definition of the instantaneous voltage

\begin{align*} \color{blue}{u_{\rm L}(t)} &= \sqrt{2}U \sin(\omega t + \varphi_u) \end{align*}

With the defining formula for inductivity, we get: \begin{align*} \color{blue}{u_{\rm L}(t)} &= L\cdot {{{\rm d}\color{red} {i_{\rm L}(t)}}\over{{\rm d}t}} \\ \rightarrow \color{red}{i_{\rm L}(t)} &= {{1}\over{L}} \int \color{blue}{u_{\rm L}(t)} {\rm d}t \\ &= - \sqrt{2} {{U}\over{\omega L}} \cos(\omega t + \varphi_u) \end{align*}

Since we assume pure AC signals the integration constant has to be 0.
This formulas lead to an instantaneous power $p_L(t)$ of

\begin{align*} p_L(t) &= \color{blue}{u(t)} \cdot \color{red}{i(t)} \\ &= - 2\cdot {{U^2}\over{\omega L}} \cdot \sin( \omega t + \varphi_u) \cos(\omega t + \varphi_u) \\ &= - {{U^2}\over{\omega L}} \cdot \sin( 2\cdot (\omega t + \varphi_u)) \\ \end{align*}

Again a trigonometric identity (Double-angle formula “$\sin(2x) = 2 \sin(x)\cos(x)$”) was used.

Again this result leads to:

1. The part $\sin( 2\cdot (\omega t + \varphi_u))$ has an average value of $0$.
2. Therefore, the average value of $p_L(t)=0$

• Instantaneous values of power at $R$, $L$, $C$
• Active, reactive, apparent, and complex power

This effect can also be seen in the following simulation: The simulation shows three loads, all with an impedance of $|Z| = 1 ~\rm k\Omega$. The diagram on top of each circuit shows the instantaneous voltage, current and power.

1. Ohmic load: The instantaneous voltage is in phase with the instantaneous current. The instantaneous power is always non-negative. The average power is $P=U^2/R = {{1}\over{2}} \hat{U}^2/R= {{1}\over{2}}(6V)^2/1 ~\rm k\Omega = 18 ~\rm mW$
2. Inductive load: The voltage is ahead of the current. The phase angle is $+90°$ (which also reflects the $+{\rm j}$ in the inductive impedance $+{\rm j}\omega L$). The instantaneous is half positive, half negative; the average power is zero (in the simulation not completely visible).
3. Capacitive load: The voltage is lagging the current. The phase angle is $-90°$ (which also reflects the $-{\rm j}$ in the capacitive impedance ${{1}\over{{\rm j}\omega C}}$). The instantaneous is again half positive, half negative; the average power is zero (in the simulation not completely visible).

arbitrary two-terminal Component

For an arbitrary component, we do not have any defining formula. But, the $u(t)$ and $i(t)$ can generally be defined as:

\begin{align*} \color{blue}{u(t)} &= \sqrt{2}U \sin(\omega t + \varphi_u) \\ \color{red }{i(t)} &= \sqrt{2}I \sin(\omega t + \varphi_i) \\ \end{align*}

This leads to an instantaneous power $p(t)$ of

\begin{align*} p(t) &= \color{blue}{u(t)} \cdot \color{red}{i(t)} \\ &= 2\cdot UI \sin(\omega t + \varphi_u) \sin(\omega t + \varphi_i) \\ \end{align*}

The formula can be further simplified with the help of the following equations

• $\varphi = \varphi_u - \varphi_i \quad \rightarrow \varphi_i = \varphi_u - \varphi$
• $ \sin(\Box - \varphi) = \sin(\Box) \cos \varphi - \cos(\Box) \sin \varphi $
• $2\sin\Box \sin\Box = 1 - \cos(2\Box)$
• $2\sin\Box \cos\Box = \sin(2\Box)$

\begin{align*} p(t) = UI\big( \cos\varphi \left( 1- \cos(2(\omega t + \varphi_u)\right) - \sin\varphi \cdot \sin(2(\omega t + \varphi_u) \big) \\ \end{align*}

This result is twofold:

1. The part $ \cos\varphi \; \cdot \; \left( 1- \cos(2(\omega t + \varphi_u)\right)$ results into a non-zero average - explicitly this part is $1$ in average. On average the first part of the formula results in $UI \cos\varphi$.
2. The part $- \sin\varphi \; \cdot \; \sin(2(\omega t + \varphi_u))$ is zero on average, so the second part of the formula results in zero. The amplitude of the second part is $UI \sin\varphi$

Based on the given formulas the three types of power are connected with each other. Since the apparent power is given by $S=U\cdot I$, the active power $P = U\cdot I \cdot \cos \varphi = S \cdot \cos \varphi $ and the reactive power $Q = S \cdot \sin \varphi $, the relationship can be shown in a triangle (see figure 2).

Generally, the apparent power can also be interpreted as a complex value:

\begin{align*} \underline{S} &= S \cdot {\rm e}^{{\rm j}\varphi} \\ &= U \cdot I \cdot {\rm e}^{{\rm j}\varphi} \end{align*}

Based on the definition of the phase angle $\varphi = \varphi_U - \varphi_I$, this can be divided into:

\begin{align*} \underline{S} &= U \cdot I \cdot {\rm e}^{ {\rm j}(\varphi_U - \varphi_I)} \\ &= \underbrace{U \cdot {\rm e}^{{\rm j}\varphi_U}}_{\underline{U}} \cdot \underbrace{I \cdot {\rm e}^{-{\rm j}\varphi_I}}_{\underline{I}^*} \end{align*}

where $\underline{I}^*$ is the complex conjungated value of $\underline{I}$.

The following simulation shows three ohmic-inductive loads, all with an impedance of $|Z| = 1 ~\rm k\Omega$, however with different phase angles $\varphi$. The diagram on top of each circuit shows the instantaneous voltage, current and power. Similar to the last simulation, a pure ohmic resistance would consume an average power of $P=U^2/R = {{1}\over{2}} \hat{U}^2/R= {{1}\over{2}}(6~\rm V)^2/1~\rm k\Omega = 18~\rm mW$. The three diagrams shall be discussed shortly.

1. Phase angle $\varphi = 10°$: Nearly all of the impedance is given by the resistance and therefore the real part of the impedance. The instantaneous voltage is nearly in phase with the current. The instantaneous power is almost always larger than zero. The average power with $17.47~\rm mW$ is about the same as for an ohmic impedance.
2. Phase angle $\varphi = 60°$: It is clearly visible, that instantaneous voltage and current are out of phase. The instantaneous power is often lower than zero. The ohmic resistor has $500 ~\rm \Omega = {{1}\over{2}}|Z|$, but does not show half of the voltage! This is because the addition has to respect the complex behavior of the values. The complex part is $90°$ perpendicular to the real part - so they generate a right-angled triangle. The average power with $9~\rm mW$ is exactly half of the power for an ohmic impedance since only the resistance provides a way for consuming power permanently.
3. Phase angle $\varphi = 84.28°$: The phase angle is calculated in such a way, that the resistance is only 10% of the amplitude of the impedance $|Z|$. In this case, the load is nearly pure inductive. The instantaneous power is consequently almost half of the time lower than zero. The average power here is also only $10%$ of the power for a pure ohmic impedance.

The next simulation enables us to play around with the phase angle of an impedance. The circuit on the left side is a bit harder to understand but consists of a resistive (real) impedance and a complex impedance, which are driven by an AC voltage source. All of these components are parameterizable in such a way that the phase angle can be manipulated by the slider on the right side.
In the middle part reflects the time course of:

• The instantaneous power $p$ of the real part (active power), the imaginary part (reactive power) and overall power.
• The instantaneous voltage and current.

On the right-hand side, the impedance Phasor is shown (lower diagram). The upper diagram depicts the $u$-$i$-diagram, which would be a perfect line for a pure ohmic resistance (since $u_R = R \cdot i_R$) and a circle for a pure complex impedance (since the phase angle of $\pm 90°$ between $u_{L, C}$ and $i_{L, C}$). The simulation is in this part not completely perfect: The pure line and circle are sometimes not reachable.

The following questions can be solved with this simulation:

1. How does the amplitude of the active and reactive instantaneous power change, when the phase angle is changed between $-90°...+90°$?
2. What is the phase shift between the active and reactive instantaneous power?

¹⁾

Also, the last simulation shows the relation between the phase angle (here: $\alpha$) and instantaneous values, like power, voltage, and current.

Applications

Power Factor Correction

Cables and components have to conduct the sum of active and reactive currents, but only the active current is used outside of the circuit. Therefore, a common goal is to minimize the reactive part. The technical way to represent this is the power factor $pf$ is used.

How does the power factor show the problematic effects? For this one can investigate the situation of an ohmic-inductive load $\underline{Z}_L$ which is connected to a voltage $\underline{U}_0$ source with a wire $R_{wire}$. This circuit is shown in figure 4.

The usable output power is $P_L = U_{\rm L} \cdot I \cdot \cos \varphi$. Based on this, the current $\underline{I}$ is:

\begin{align*} I = {{P_L}\over{U_{\rm L} \cdot \cos \varphi}} \end{align*}

The power loss of the wire $P_{wire}$ is therefore:

\begin{align*} P_{\rm wire} &= R_{\rm wire} \cdot I^2 \\ &= R_{\rm wire} \cdot {{P_L^2}\over{U_{\rm L}^2 \cdot \cos^2 \varphi}} \end{align*}

This means: As smaller, the power factor $\cos \varphi$, as more power losses $P_{\rm wire}$ will be generated. More power losses $P_{\rm wire}$ lead to more heat up to or even beyond the maximum temperature. To compensate for this, the cross-section of the wire has to be increased, which means more copper.

Alternatively, a bad power factor can be compensated with a counteracting complex impedance. This compensating impedance has to provide enough power with the opposite sign to cancel out the unwanted reactive power. The following simulation shows an uncompensated circuit and a circuit with power factor correction. In the ladder, the voltage on the load resistor is the same, but the current provided by the power supply is smaller.

Another explanation of the power factor can be seen here:

Impedance matching

not covered in this course

Common pitfalls

• …

Exercises

Worked examples

Exercise 7.1.1 Power and Power Factor I

A passive component is fed by a sinusoidal AC voltage with the RMS value $U=230~\rm V$ and $f=50.0~\rm Hz$. The RMS current on this component is $I=5.00~\rm A$ with a phase angle of $\varphi=+60°$.

1. Draw the equivalent circuits based on a series and a parallel circuit of two components.

Result

2. Calculate the equivalent components for both circuits.

Solution

The apparent impedance is: \begin{align*} Z = |\underline{Z}| &={{U}\over{I}}= {{230~\rm V}\over{5~\rm A}} = 46 ~\Omega \\ \end{align*}

For the series circuit, the impedances add up like: $R_s + {\rm j}\cdot X_{Ls} = \underline{Z} $, and $R_s = |\underline{Z}| \cos\varphi$ such as $X_{Ls} = |\underline{Z}| \sin\varphi$. Therefore: \begin{align*} R_s &=&{{U}\over{I}} \cdot \cos \varphi &=& {{230~\rm V}\over{5.00~\rm A}} \cdot \cos 60° &=& \boldsymbol{23.0 ~\Omega} & \\ X_{Ls} &=&{{U}\over{I}} \cdot \sin \varphi &=& {{230~\rm V}\over{5.00~\rm A}} \cdot \sin 60° &=& 39.8 ~\Omega &= \omega \cdot L_s \\ \rightarrow L_s &=& {{X_{Ls}}\over{2\pi f}} &=& {{{{U}\over{I}} \cdot \sin \varphi}\over{2\pi f}} &=& \boldsymbol{127~\rm mH} \end{align*}

For the parallel circuit, the impedances add up like ${{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}}= {{1}\over{\underline{Z}}} $ with $\underline{Z} = {{U}\over{I}}\cdot e^{j\cdot \varphi}$.

There are multiple ways to solve this problem. Two ways shall be shown here:

with the Euler representation

Given the formula $\underline{Z} = {{U}\over{I}}\cdot e^{j\cdot \varphi}$ the following can be derived: \begin{align*} {{1}\over{\underline{Z}^{\phantom{A}}}} &= {{I}\over{U}}\cdot e^{-j\cdot \varphi} \\ &= {{1}\over{Z}}\cdot e^{-j\cdot \varphi} &&= {{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}} \\ &= {{1}\over{Z}}\cdot \left( \cos(\varphi) - {\rm j}\cdot \sin(\varphi) \right) &&= {{1}\over{R_p}} - {{\rm j}\over{X_{Lp}}} \\ \end{align*}

Therefore, the following can be concluded: \begin{align*} {{1}\over{Z}}\cdot \cos(\varphi) &= {{1}\over{R_p}} &&\rightarrow && R_p &= {{Z}\over{\cos(\varphi)}} \\ - {\rm j}\cdot \sin(\varphi) &= - {{\rm j}\over{ X_{Lp}}} &&\rightarrow && X_{Lp} &= {{Z}\over{\sin(\varphi)}} \\ \end{align*}

with the calculated values of the series circuit

Another way is to use the formulas of $R_s$ and $X_{Ls}$ from before.

\begin{align*} {{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}} &=& {{1}\over{R_s + {\rm j}\cdot X_{Ls}}} \\ {{1}\over{R_p}} - {\rm j} {{1}\over{X_{Lp}}} &=& {{R_s - {\rm j}\cdot X_{Ls}}\over{R_s^2 + X_{Ls}^2}} \\ &=& {{Z \cdot \cos \varphi - {\rm j}\cdot Z \cdot \sin \varphi }\over{Z^2}} \\ &=& { {\cos \varphi - {\rm j} \cdot \sin \varphi } \over{Z}} \\ \end{align*}

Therefore

Now, the real and imaginary part is analyzed individually. First the real part:

\begin{align*} {{1}\over{R_p}} &=& {{\cos \varphi}\over{Z}} \\ \rightarrow R_p &=& {{Z}\over{\cos \varphi}} &=& {{46 ~\Omega}\over{\cos 60°}} \end{align*}

\begin{align*} {{1}\over{X_{Lp}}} &= {{\sin \varphi}\over{Z}} & \\ \rightarrow X_{Lp} &= {{Z}\over{\sin \varphi}} = {{46 ~\Omega}\over{\sin 60°}} \\ \rightarrow L_p &= {{46 ~\Omega}\over{2\pi \cdot 50~\rm Hz \cdot \sin 60°}} \end{align*}

Result

For the series circuit: \begin{align*} R_s &= {23 ~\Omega} \\ L_s &= {127 ~\rm mH} \\ \end{align*}

For the parallel circuit: \begin{align*} R_p &= {92 ~\Omega} \\ L_p &= {169 ~\rm mH} \\ \end{align*}

3. Calculate the real, reactive, and apparent power based on the equivalent components for both circuits from 2. .

Solution

The general formula for the apparent power is $\underline{S} = U \cdot I \cdot e^{\rm j\varphi}$.
By this, the following can be derived: \begin{align*} \underline{S} &= U \cdot I \cdot e^{\rm j\varphi} \\ &= Z \cdot I^2 \cdot e^{\rm j\varphi} &&= \underline{Z} \cdot I^2 \\ &= {{U^2}\over{Z}} \cdot e^{\rm j\varphi} &&= {{U^2}\over{\underline{Z}^{*\phantom{I}}}} \\ \end{align*}

These formulas are handy for both types of circuits to separate the apparent power into real part (real power) and complex part (apparent power):

1. for series circuit: $\underline{S} =\underline{Z} \cdot I^2 $ with $\underline{Z} = R + {\rm j} X_L$
2. for parallel circuit: $\underline{S} ={{U^2}\over{\underline{Z}^{*\phantom{I}}}} $ with ${{1} \over {\underline{Z}^{\phantom{I}}} } = {{1}\over{R}} + {{1}\over{{\rm j} X_L}} \rightarrow {{1} \over {\underline{Z}^{*\phantom{I}}} } = {{1}\over{R}} + {{\rm j}\over{ X_L}} $

Therefore:

	series circuit	parallel circuit
active power	\begin{align*} P_s &= R_s \cdot I^2 \\ &= 23.0 ~\Omega \cdot (5{~\rm A})^2 \\ &= 575 {~\rm W} \end{align*}	\begin{align*} P_p &= {{U_p^2}\over{R_p}} \\ &= {{(230{~\rm V})^2}\over{92~\Omega}} = 575 {~\rm W} \end{align*}
reactive power	\begin{align*} Q_s &= X_{Ls} \cdot I^2 \\ &= 39.8 ~\Omega \cdot (5{~\rm A})^2 \\ &= 996 {~\rm Var} \end{align*}	\begin{align*} Q_p &= {{U_p^2}\over{X_{Lp}}} \\ &= {{(230{~\rm V})^2}\over{53.1 ~\Omega}} = 996 {~\rm Var} \end{align*}
apparent power	\begin{align*} S_s &= \sqrt{P_s^2 - Q_s^2} \\ &= I^2 \cdot \sqrt{R_s^2 + X_{Ls}^2} \\ &= 1150 {~\rm VA} \end{align*}	\begin{align*} S_p &= \sqrt{P_s^2 - Q_s^2} \\ &= U^2 \cdot \sqrt{{{1}\over{R_p^2}} + {{1}\over{X_{Lp}^2}}} \\ &= 1150 {~\rm VA} \end{align*}

4. Check the solutions from 3. via direct calculation based on the input in the task above.

Solution

active power: \begin{align*} P &= U \cdot I \cdot \cos \varphi \\ &= 220{~\rm V} \cdot 5{~\rm A} \cos 60° \\ &= 575 {~\rm W} \end{align*}

reactive power: \begin{align*} Q &= U \cdot I \cdot \sin \varphi \\ &= 230{~\rm V} \cdot 5{~\rm A} \sin 60° \\ &= 996 {~\rm Var} \end{align*}

apparent power: \begin{align*} S &= U \cdot I \\ &= 230{~\rm V} \cdot 5{~\rm A} \\ &= 1150 {~\rm VA} \end{align*}

Exercise 7.1.2 Power and Power Factor II

A magnetic coil shows at a frequency of $f=50.0 {~\rm Hz}$ the voltage of $U=115{~\rm V}$ and the current $I=2.60{~\rm A}$ with a power factor of $\cos \varphi = 0.30$

1. Calculate the real power, the reactive power, and the apparent power.
2. Draw the equivalent parallel circuit. Calculate the active and reactive part of the current.
3. Draw the equivalent series circuit. Calculate the ohmic and inductive impedance and the value of the inductivity.

Result for 1.

The real power is \begin{align*} P &= U \cdot I \cdot \cos \varphi \\ &= 115{~\rm V} \cdot 2.6{~\rm A} \cdot 0.3 \\ &= 89.7 {~\rm W} \end{align*}

The reactive power is \begin{align*} Q &= U \cdot I \cdot \sin \varphi \\ &= 115{~\rm V} \cdot 2.6{~\rm V} \cdot \sqrt{1 - 0.3^2} \\ &= 285 {~\rm Var} \end{align*}

The apparent power is \begin{align*} S &= U \cdot I \\ &= 115{~\rm V} \cdot 2.6{~\rm A} \\ &= 299 {~\rm VA} \end{align*}

Result for 2.

The complex current $\underline{I}$ is given as:

\begin{align*} \underline{I} &= I_R + {\rm j} \cdot I_L \\ &= I \cdot \cos\varphi - {\rm j} \cdot I \cdot \sin\varphi \end{align*}

The active and reactive part of the current is therefore: \begin{align*} I_R &= 2.60{~\rm A} \cdot 0.30 &= 0.78 {~\rm A} \\ I_L &= - 2.60{~\rm A} \cdot \sqrt{1 - 0.30^2} &= 2.48 {~\rm A} \end{align*}

Result for 3.

Important: The cosine function is ambiguous! Based on $\cos \varphi = 0.30$ it is unclear, whether $\varphi$ is positive or negative.
Therefore, only based on the power factor it is unclear whether the circuit is ohmic-inductive or ohmic-capacitive! However, this is explicitly given in the problem definition.

\begin{align*} Z_s &= {{U}\over{I}} &=& {{115{~\rm V}}\over{2.60{~\rm A}}} &=& 44.2 ~\Omega \\ R_s &= {{U}\over{I}} \cdot \cos \varphi &=& {{115{~\rm V}}\over{2.60{~\rm A}}} \cdot 0.30 &=& 13.3 ~\Omega \\ X_{Ls} &= {{U}\over{I}} \cdot \sin \varphi &=& {{115{~\rm V}}\over{2.60{~\rm A}}} \cdot \sqrt{1 - 0.30^2} &=& 42.2 ~\Omega \\ \\ L_s &= 134~\rm mH \end{align*}

Embedded resources