Question

assume that the source voltage is given as: $v_{in}(t) = 25cos(4000t) , \text{v}$
what is the phasor current $i_0$?
$i_0 = 0.625angle -1.875^circ , \text{a}$
$i_0 = 1.98angle -71.56^circ , \text{a}$
$i_0 = 19.76angle 18.43^circ , \text{a}$
$i_0 = 1.98angle -1.25^circ , \text{a}$
$i_0 = 2.5angle -90^circ , \text{a}$

Explanation:

Step1: Calculate reactances

First, find the inductive reactance $X_L = \omega L$ and capacitive reactance $X_C = \frac{1}{\omega C}$, where $\omega=4000$ rad/s:
$X_L = 4000 \times 2.5 \times 10^{-3} = 10\ \Omega$
$X_C = \frac{1}{4000 \times 12.5 \times 10^{-6}} = 20\ \Omega$
Convert components to phasors:

• Inductor: $jX_L = j10\ \Omega$
• Capacitor: $-jX_C = -j20\ \Omega$
• Source phasor: $V_{in} = 25\angle0^\circ\ \text{V}$

Step2: Simplify rightmost parallel branch

Combine the $20\ \Omega$ resistor and $-j20\ \Omega$ capacitor in parallel:
$$Z_1 = \frac{20(-j20)}{20 - j20} = \frac{-j400}{20(1-j)} = \frac{-j20}{1-j} = \frac{-j20(1+j)}{(1-j)(1+j)} = \frac{20 - j20}{2} = 10 - j10\ \Omega$$

Step3: Add series 10Ω resistor

$$Z_2 = 10 + Z_1 = 10 + 10 - j10 = 20 - j10\ \Omega$$

Step4: Parallel with inductor $j10\ \Omega$

This is the load parallel to the 5Ω resistor:
$$Z_3 = \frac{j10(20 - j10)}{j10 + 20 - j10} = \frac{j200 - j^2100}{20} = \frac{100 + j200}{20} = 5 + j10\ \Omega$$

Step5: Total circuit impedance

Add the series 5Ω resistor:
$$Z_{total} = 5 + Z_3 = 5 + 5 + j10 = 10 + j10\ \Omega$$

Step6: Total source current $I_{total}$

$$I_{total} = \frac{V_{in}}{Z_{total}} = \frac{25\angle0^\circ}{10 + j10} = \frac{25}{10\sqrt{2}\angle45^\circ} = \frac{2.5}{\sqrt{2}}\angle-45^\circ \approx 1.7678\angle-45^\circ\ \text{A}$$

Step7: Use current division for $I_0$

Current divides between $j10\ \Omega$ and $Z_2=20-j10\ \Omega$:
$$I_0 = I_{total} \times \frac{Z_2}{j10 + Z_2}$$
Substitute values:
$$I_0 = 1.7678\angle-45^\circ \times \frac{20 - j10}{20}$$
First, $20 - j10 = 22.3607\angle-26.565^\circ$, so:
$$I_0 = 1.7678\angle-45^\circ \times \frac{22.3607\angle-26.565^\circ}{20}$$
$$I_0 = 1.7678 \times 1.1180\angle(-45^\circ-26.565^\circ) \approx 1.976\angle-71.565^\circ\ \text{A}$$

Answer:

$I_0 = 1.98\angle - 71.56^\circ$ A