An impedance Z₁(2 – j5) Ω is connected in parallel with another impedance of Z₂(1 + j1) Ω. If the applied voltage is 17∠0° V, then the currents through Z₁ and Z₂ are respectively

Question

Prepp Team · Accepted Answer

Calculating Currents in a Parallel AC Circuit with Impedances

This problem involves finding the individual currents flowing through two different impedances connected in parallel when a voltage is applied across them. In a parallel circuit, the voltage across each branch is the same as the total applied voltage. Therefore, we can use Ohm's Law directly for each impedance to find the current through it.

The given values are:

Impedance 1: $Z_1 = (2 - j5) \, \Omega$
Impedance 2: $Z_2 = (1 + j1) \, \Omega$
Applied Voltage: $V = 17\angle0^\circ \, V$

Ohm's Law for AC circuits states that current ($I$) is equal to voltage ($V$) divided by impedance ($Z$), represented as $I = V/Z$. To perform this calculation with complex impedances and phasor voltages, it is often easiest to work with the values in polar form.

Step-by-Step Calculation of Current Through Z₁ (I₁)

First, we need to calculate the current $I_1$ flowing through impedance $Z_1$.

Given $V = 17\angle0^\circ \, V$ and $Z_1 = (2 - j5) \, \Omega$.

Convert $Z_1$ from rectangular form to polar form ($|Z_1|\angle\theta_1$):

Magnitude $|Z_1| = \sqrt{(\text{Real part})^2 + (\text{Imaginary part})^2}$
Angle $\theta_1 = \arctan\left(\frac{\text{Imaginary part}}{\text{Real part}}\right)$

For $Z_1 = (2 - j5) \, \Omega$:

$|Z_1| = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}$
$|Z_1| \approx 5.385 \, \Omega$
$\theta_1 = \arctan\left(\frac{-5}{2}\right) = \arctan(-2.5)$
$\theta_1 \approx -68.198^\circ$

So, $Z_1 \approx 5.385\angle-68.2^\circ \, \Omega$.

Now, calculate $I_1$ using Ohm's Law $I_1 = V/Z_1$:

In polar form, division is done by dividing the magnitudes and subtracting the angles:

$I_1 = \frac{|V|\angle\angle V}{|Z_1|\angle\theta_1} = \frac{|V|}{|Z_1|}\angle(\angle V - \theta_1)$

$I_1 = \frac{17}{5.385}\angle(0^\circ - (-68.2^\circ))$

$|I_1| \approx 3.157$

$\angle I_1 = 68.2^\circ$

Therefore, $I_1 \approx 3.16\angle68.2^\circ \, A$.

Step-by-Step Calculation of Current Through Z₂ (I₂)

Next, we calculate the current $I_2$ flowing through impedance $Z_2$.

Given $V = 17\angle0^\circ \, V$ and $Z_2 = (1 + j1) \, \Omega$.

Convert $Z_2$ from rectangular form to polar form ($|Z_2|\angle\theta_2$):

Magnitude $|Z_2| = \sqrt{(\text{Real part})^2 + (\text{Imaginary part})^2}$
Angle $\theta_2 = \arctan\left(\frac{\text{Imaginary part}}{\text{Real part}}\right)$

For $Z_2 = (1 + j1) \, \Omega$:

$|Z_2| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}$
$|Z_2| \approx 1.414 \, \Omega$
$\theta_2 = \arctan\left(\frac{1}{1}\right) = \arctan(1)$
$\theta_2 = 45^\circ$

So, $Z_2 \approx 1.414\angle45^\circ \, \Omega$.

Now, calculate $I_2$ using Ohm's Law $I_2 = V/Z_2$:

$I_2 = \frac{|V|\angle\angle V}{|Z_2|\angle\theta_2} = \frac{|V|}{|Z_2|}\angle(\angle V - \theta_2)$

$I_2 = \frac{17}{1.414}\angle(0^\circ - 45^\circ)$

$|I_2| \approx 12.022$

$\angle I_2 = -45^\circ$

Therefore, $I_2 \approx 12.02\angle-45^\circ \, A$.

Summary of Calculated Currents

Based on the calculations, the currents through $Z_1$ and $Z_2$ are approximately:

$I_1 \approx 3.16\angle68.2^\circ \, A$
$I_2 \approx 12.02\angle-45^\circ \, A$

Comparing these values with the given options, we find the option that matches these results.

Option	Currents	Match?
1	3.16∠68.2° A and 10.02 ∠-35° A	$I_1$ matches, $I_2$ does not
2	2.16∠48.2° A and 10.02 ∠-35° A	Neither matches
3	3.16∠68.2° A and 12.02 ∠-45° A	Both match
4	2.16∠48.2° A and 12.02 ∠-45° A	$I_2$ matches, $I_1$ does not

The calculated currents $I_1 \approx 3.16\angle68.2^\circ \, A$ and $I_2 \approx 12.02\angle-45^\circ \, A$ match the values given in option 3.

Revision Table: AC Parallel Circuit Analysis

Concept	Description	Formula/Method
Parallel Circuit Voltage	Voltage is the same across all parallel branches.	$V_{total} = V_1 = V_2 = \dots$
Ohm's Law (AC)	Relates voltage, current, and impedance.	$V = IZ$ or $I = V/Z$ or $Z = V/I$
Rectangular Form Impedance	Represents resistance (real part) and reactance (imaginary part).	$Z = R + jX$
Polar Form Impedance	Represents magnitude and phase angle. Useful for multiplication and division.	$Z = \|Z\|\angle\theta$
Conversion (Rectangular to Polar)	Magnitude and angle from real and imaginary parts.	$\|Z\| = \sqrt{R^2 + X^2}$, $\theta = \arctan(X/R)$
Phasor Division	Divide magnitudes, subtract angles.	$\frac{\|V\|\angle\angle V}{\|Z\|\angle\theta} = \frac{\|V\|}{\|Z\|}\angle(\angle V - \theta)$

Additional Information: AC Circuit Impedance and Phasors

In AC circuits, components like resistors, inductors, and capacitors oppose the flow of alternating current. This opposition is called impedance ($Z$). Unlike resistance in DC circuits, impedance is a complex quantity because inductors and capacitors cause a phase shift between the voltage and current.

Resistor (R): Impedance is just the resistance, $Z_R = R\angle0^\circ = R$.
Inductor (L): Impedance is $Z_L = j\omega L = \omega L\angle90^\circ$. $\omega = 2\pi f$ is the angular frequency.
Capacitor (C): Impedance is $Z_C = \frac{1}{j\omega C} = \frac{-j}{\omega C} = \frac{1}{\omega C}\angle-90^\circ$.

Complex numbers are used to represent impedance, voltage, and current in AC circuits using phasors. A phasor is a rotating vector that represents a sinusoidal quantity. Using complex numbers (rectangular form $a+jb$ or polar form $|Z|\angle\theta$) simplifies calculations involving phase shifts caused by reactances (the imaginary part of impedance).

When impedances are in parallel, the total impedance ($Z_{total}$) is calculated using a reciprocal formula similar to parallel resistances, but with complex numbers: $\frac{1}{Z_{total}} = \frac{1}{Z_1} + \frac{1}{Z_2} + \dots$ The total current is $I_{total} = V/Z_{total}$, and individual currents in parallel branches are found by $I_k = V/Z_k$, as demonstrated in this problem.