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11 \mathrm{~V}, 3 \mathrm{kHz}

single-phase ac supply feeds a

90 \Omega

r...

May 20, 2024

Solution

The real part of the complex impedance is simply the resistance of the resistor. Therefore, the real part of the impedance is

R = 90 \Omega

The imaginary part of the complex impedance is due to the capacitor. The impedance of a capacitor is given by

Z_C = -\frac{1}{j\omega C}

where

\omega = 2\pi f

. Given

f = 3 \text{kHz}

and

C = 3 \mu \text{F}

, we have:

\omega = 2\pi \times 3000 \text{Hz} = 6000\pi \text{rad/s}

Z_C = -\frac{1}{j \times 6000\pi \times 3 \times 10^{-6}} = -\frac{1}{j \times 0.01885} = -j53.05 \Omega

So, the imaginary part of the impedance is

-53.05 \Omega

The real part of the current can be calculated using Ohm's Law. The total impedance is

Z = 90 - j53.05 \Omega

. The magnitude of the impedance is:

|Z| = \sqrt{90^2 + (-53.05)^2} = \sqrt{8100 + 2814.0025} = \sqrt{10914.0025} \approx 104.5 \Omega

The voltage is given as

11 \text{V}

, so the current is:

I = \frac{V}{|Z|} = \frac{11}{104.5} \approx 0.105 \text{A} = 105 \text{mA}

The real part of the current is:

I_R = I \cos(\theta)

where

\theta = \tan^{-1}\left(\frac{-53.05}{90}\right) \approx -30.96^\circ

I_R = 105 \cos(-30.96^\circ) \approx 105 \times 0.857 = 90 \text{mA}

The imaginary part of the current is:

I_I = I \sin(\theta)

I_I = 105 \sin(-30.96^\circ) \approx 105 \times -0.515 = -54 \text{mA}

The magnitude of the current is already calculated in part (c) as:

|I| = 105 \text{mA}

Answer

The real part of the impedance is 90 Ω. The imaginary part of the impedance is -53.05 Ω. The real part of the current is 90 mA. The imaginary part of the current is -54 mA. The magnitude of the current is 105 mA.

Key Concept

Impedance in AC circuits: Impedance is a complex quantity that combines resistance (real part) and reactance (imaginary part). The impedance of a resistor is purely real, while the impedance of a capacitor is purely imaginary and negative.

Explanation

By calculating the impedance of the resistor and capacitor in series, we can determine the real and imaginary parts of the total impedance. Using Ohm's Law, we can then find the real and imaginary parts of the current.