Question Three Two impedances, $$ Z_{1}=(3+j 6) \Omega $$ and $$ Z_{2}=(4-j 3) \Omega $$ are connected in series to a supply voltage of 120 V . Determine the magnitude of the current and phase angle relative the voltage.

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We are given two impedances in series:
$$
Z_1 = 3 + j6\ \Omega,\quad Z_2 = 4 - j3\ \Omega.
$$

**Step 1: Compute the Total Impedance**

Since the impedances are in series, we add them:
$$
Z_{\text{total}} = Z_1 + Z_2 = (3 + j6) + (4 - j3) = (3+4) + j(6-3) = 7 + j3\ \Omega.
$$

**Step 2: Find the Magnitude and Phase of $$Z_{\text{total}}$$**

Magnitude:
$$
\lvert Z_{\text{total}} \rvert = \sqrt{7^2 + 3^2} = \sqrt{49 + 9} = \sqrt{58}.
$$

Phase angle:
$$
\theta = \arctan\left(\frac{3}{7}\right).
$$
Approximating:
$$
\theta \approx \arctan(0.4286) \approx 23.2^\circ.
$$

**Step 3: Calculate the Current**

The supply voltage is $$ V = 120\ \text{V} $$ (assumed to be the reference phasor with $$0^\circ$$ phase). Using Ohm's law for AC circuits,
$$
I = \frac{V}{Z_{\text{total}}}.
$$
The magnitude of the current is:
$$
\lvert I \rvert = \frac{120}{\lvert Z_{\text{total}} \rvert} = \frac{120}{\sqrt{58}}.
$$
Approximating, since $$\sqrt{58}\approx 7.62$$,
$$
\lvert I \rvert \approx \frac{120}{7.62} \approx 15.75\ \text{A}.
$$

The phase of the current is:
$$
\text{Angle of } I = 0^\circ - \theta = -23.2^\circ.
$$
This negative angle indicates that the current lags behind the voltage.

**Final Answer:**

The magnitude of the current is approximately $$15.75\ \text{A}$$ and its phase angle is $$-23.2^\circ$$ relative to the voltage.
The current is approximately 15.75 A and lags the voltage by about 23.2 degrees.

Question Three Two impedances, \( Z_{1}=(3+j 6) \Omega \) and \( Z_{2}=(4-j 3) \Omega \) are connected in series to a supply voltage of 120 V . Determine the magnitude of the current and phase angle relative the voltage.

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