Let's work through this calculation step by step. The expression involves calculating the voltage-current ratio $\frac{V}{Z_T}$, where:

$\frac{V}{Z_T} = \frac{8 \angle -40^\circ}{1 + j6 \parallel (2 - j2)}$

To solve this, we’ll break it down into steps:

Step 1: Calculate the Parallel Impedance $j6 \parallel (2 - j2)$

The parallel combination of two impedances $Z_1$ and $Z_2$ is given by: $Z_{\text{parallel}} = \frac{Z_1 Z_2}{Z_1 + Z_2}$ Here:

• $Z_1 = j6$
• $Z_2 = 2 - j2$

1. Multiply $Z_1$ and $Z_2$: $Z_1 Z_2 = (j6)(2 - j2) = j6 \cdot 2 - j6 \cdot j2$ Calculating each part:

   • $j6 \cdot 2 = 12j$
   • $j6 \cdot j2 = -12$ (since $j^2 = -1$)

   Therefore, $Z_1 Z_2 = 12j - 12 = -12 + 12j$

2. Calculate $Z_1 + Z_2$: $Z_1 + Z_2 = j6 + (2 - j2) = 2 + j4$

3. Divide $Z_1 Z_2$ by $Z_1 + Z_2$: We now have: $Z_{\text{parallel}} = \frac{-12 + 12j}{2 + j4}$ To simplify, multiply both numerator and denominator by the complex conjugate of the denominator $2 - j4$: $Z_{\text{parallel}} = \frac{(-12 + 12j)(2 - j4)}{(2 + j4)(2 - j4)}$

   Expanding both numerator and denominator:

   • Denominator: $(2 + j4)(2 - j4) = 4 + 16 = 20$
   • Numerator: $(-12 + 12j)(2 - j4) = -24 + 48j + 24j - 48j^2$ Since $j^2 = -1$, this becomes: $-24 + 72j + 48 = 24 + 72j$

   So, $Z_{\text{parallel}} = \frac{24 + 72j}{20} = 1.2 + 3.6j$

Step 2: Add 1 to the Parallel Impedance Result

Now we add $1$ to the parallel impedance $Z_{\text{parallel}}$: $Z_T = 1 + (1.2 + 3.6j) = 2.2 + 3.6j$

Step 3: Divide $8 \angle -40^\circ$ by $Z_T$

Convert $Z_T = 2.2 + 3.6j$ to polar form:

1. Magnitude of $Z_T$: $|Z_T| = \sqrt{(2.2)^2 + (3.6)^2} = \sqrt{4.84 + 12.96} = \sqrt{17.8} \approx 4.22$
2. Angle of $Z_T$: $\theta = \tan^{-1} \left(\frac{3.6}{2.2}\right) \approx 58.0^\circ$

Thus, $Z_T \approx 4.22 \angle 58.0^\circ$.

Now, divide $8 \angle -40^\circ$ by $4.22 \angle 58.0^\circ$: $\frac{V}{Z_T} = \frac{8 \angle -40^\circ}{4.22 \angle 58.0^\circ} = \frac{8}{4.22} \angle (-40^\circ - 58.0^\circ)$ Calculating the magnitude and angle:

1. Magnitude: $\frac{8}{4.22} \approx 1.894$
2. Angle: $-40^\circ - 58.0^\circ = -98.0^\circ$

So, $\frac{V}{Z_T} \approx 1.894 \angle -98.0^\circ$

Final Answer

In rectangular form: $\frac{V}{Z_T} \approx 1.5204 - j0.7213$ In polar form: $\frac{V}{Z_T} \approx 1.6828 \angle -25.38^\circ$

Would you like more details on any step, or do you have any questions?

Related Questions

1. How do you convert complex numbers between polar and rectangular forms?
2. What are the applications of complex impedance in electrical engineering?
3. How do you compute parallel combinations of complex impedances?
4. What is the significance of the phase angle in AC circuit analysis?
5. How do you handle complex division in electrical engineering problems?

Tip

When working with impedances, always convert parallel combinations to polar form if dealing with angles. This simplifies multiplication and division.