To solve the problem based on the provided circuit and the given parameters:

Given:

1. Phase voltage $U = 100 \, \text{V}$
2. Impedance $Z_1 = 8 + j6 \, \Omega$
3. Impedance $Z_2 = j8 \, \Omega$

We are tasked with finding:

1. The current $I$
2. The total active power provided by the three-phase source

Step-by-Step Solution:

1. Phase Current Calculation:

The circuit is a balanced three-phase system. In a balanced system, the calculation for a single phase can be applied to all three phases.

The equivalent impedance per phase is the parallel combination of $Z_1$ and $Z_2$: $Z_{\text{eq}} = \frac{Z_1 Z_2}{Z_1 + Z_2}$

Using $Z_1 = 8 + j6$ and $Z_2 = j8$:

• First, calculate $Z_1 + Z_2$: $Z_1 + Z_2 = (8 + j6) + j8 = 8 + j14$

• Next, calculate $Z_1 Z_2$: $Z_1 Z_2 = (8 + j6)(j8) = 8j8 + j6j8 = j64 - 48 = -48 + j64$

• Now, divide $Z_1 Z_2$ by $Z_1 + Z_2$: $Z_{\text{eq}} = \frac{-48 + j64}{8 + j14}$

Rationalizing the denominator: $Z_{\text{eq}} = \frac{(-48 + j64)(8 - j14)}{(8 + j14)(8 - j14)}$

Denominator: $(8 + j14)(8 - j14) = 8^2 + 14^2 = 64 + 196 = 260$

Numerator: $(-48 + j64)(8 - j14) = (-48)(8) + (-48)(-j14) + (j64)(8) + (j64)(-j14)$ $= -384 + j672 + j512 - 896$ $= -384 - 896 + j(672 + 512) = -1280 + j1184$

So: $Z_{\text{eq}} = \frac{-1280 + j1184}{260} = -4.923 + j4.554 \, \Omega$

Magnitude of $Z_{\text{eq}}$: $|Z_{\text{eq}}| = \sqrt{(-4.923)^2 + (4.554)^2} = \sqrt{24.23 + 20.74} = \sqrt{44.97} \approx 6.71 \, \Omega$

2. Phase Current $I_{\text{ph}}$:

Using $I_{\text{ph}} = \frac{U}{Z_{\text{eq}}}$: $I_{\text{ph}} = \frac{100}{6.71} \approx 14.9 \, \text{A (magnitude)}$

3. Total Active Power ($P$):

The total active power in a three-phase system is: $P = 3 \cdot U_{\text{ph}} \cdot I_{\text{ph}} \cdot \cos\phi$

Where $\cos\phi$ is the power factor, given by: $\cos\phi = \frac{\text{Re}(Z_{\text{eq}})}{|Z_{\text{eq}}|}$ $\cos\phi = \frac{-4.923}{6.71} \approx 0.735$

Substitute the values: $P = 3 \cdot 100 \cdot 14.9 \cdot 0.735$ $P \approx 3278.6 \, \text{W}$

Final Answers:

1. Phase Current $I_{\text{ph}}$: $14.9 \, \text{A}$ (magnitude)
2. Total Active Power ($P$): $3278.6 \, \text{W}$

Let me know if you need further clarification or details!

Follow-Up Questions:

1. How is impedance calculated for components in parallel?
2. What is the relationship between active, reactive, and apparent power in AC systems?
3. How does the power factor affect the efficiency of power delivery?
4. Why do balanced three-phase systems simplify power calculations?
5. What are the differences between single-phase and three-phase power systems?

Tip:

In AC circuit analysis, always calculate the power factor to understand the efficiency of power transfer.