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A voltage \(V=(150+j 180) V\) is applied across an impedance and the current is found to be \(I=(5-j 4)\) A. Determine (i) scalar impedance, (ii) reactance, and (iii) power consumed. \([36.6 \Omega, 36.6 \Omega, 30 \mathrm{~W}]\)

Short Answer

Expert verified
From our calculations, the scalar impedance is 46.6 \(\Omega\), the reactance is 36 \(\Omega\), and the power consumed is 30 W.

Step by step solution

01

Calculation of Impedance

Firstly, we need to compute the impedance \(Z\) of the circuit. Since we know the current \(I\) and voltage \(V\) across the circuit, we can use Ohm's law which for complex numbers is \(Z = V/I\). Here, \(V = 150+j180\) V and \(I = 5-j4\) A. Performing \(V/I\), we get \(Z=(30+j36) \Omega\). The scalar impedance is the magnitude of \(Z\), which can be found using \(|Z| = \sqrt{Re(Z)^2 + Im(Z)^2}\). After substituting the given values, we get \(|Z| = \sqrt{30^2 + 36^2} = 46.6 \Omega\).
02

Calculation of Reactance

Next, we will find the reactance \(X\) of the impedance. The reactance is the imaginary part of the complex impedance \(Z\). Thus, \(X = Im(Z)\). In our case, \(X = 36 \Omega\).
03

Calculation of Power Consumed

Lastly, we will calculate the power \(P\) consumed by the circuit. The power in an AC circuit is given by \(P=Re(VI*)\), where \(I*\) is the complex conjugate of \(I\). So in our case, \(I* = 5+j4 A\). Substituting the values in the equation we get \(P = Re[(150+j180)(5+j4)], and after performing the multiplication and taking the real part, we obtain \(P = 30 W\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law
Ohm's Law is a fundamental principle often used in analyzing electrical circuits. For direct current (DC) circuits, it simply states that the voltage (\( V \) ) across a resistor is equal to the product of the current (\( I \) ) through the resistor and the resistance (\( R \) ) of the resistor: \[ V = I imes R \] However, in alternating current (AC) circuits, we must consider the impedance (\( Z \) ) instead of resistance because AC circuits can have components that store energy, like capacitors and inductors. Thus, the complex form of Ohm's Law becomes:\[ V = I imes Z \]Here, \( Z \) is a complex number that includes both resistance and reactance. Reactance reflects an opposition to changes in current and is crucial in AC contexts. To find impedance, we divide the voltage by the current in the circuit: \[ Z = \frac{V}{I} \] The resulting impedance tells us how much the circuit resists the flow of AC. The magnitude \( |Z| \) of the complex impedance is the scalar impedance, calculated using the formula:\[ |Z| = \sqrt{Re(Z)^2 + Im(Z)^2} \]where \( Re(Z) \) and \( Im(Z) \) are the real and imaginary parts, respectively. Understanding impedance through its real and imaginary components helps in designing and troubleshooting AC circuits.
Reactance
Reactance in AC circuits is an essential aspect to appreciate how different types of circuit components affect current flow. Not to be confused with resistance, which applies to purely resistive circuits, reactance results from the presence of capacitors and inductors which affect the circuit differently based on frequency changes.
  • Inductive Reactance (\( X_L \) ): Occurs in a coil or inductor, where it resists changes in current. It's calculated by \( X_L = 2\pi f L \), with \( f \) being the frequency and \( L \) the inductance.
  • Capacitive Reactance (\( X_C \) ): Occurs in capacitors, which resist changes in voltage. It's given by \( X_C = \frac{1}{2\pi f C} \) where \( C \) is the capacitance.
In a complex impedance \( Z = R + jX \), reactance (\( X \) ) is the imaginary part, denoted as \( Im(Z) \). The sign of the reactance indicates whether the reactive component is inductive (positive) or capacitive (negative). Knowing this helps in understanding phase differences between voltage and current, as well as harmonic behavior in circuits.
AC Power Calculation
Calculation of power in AC circuits involves more complexity than DC circuits due to the phase difference between voltage and current. In an AC system, power can be described in three components:
  • Real Power (\( P \) ), measured in watts (\( W \) ), is the actual power consumed by the circuit to perform work.
  • Reactive Power (\( Q \) ), measured in volt-amperes reactive (\( VAR \) ), is the power oscillating back and forth due to the reactance.
  • Apparent Power (\( S \) ), measured in volt-amperes (\( VA \) ), combines real and reactive power.
To calculate real power in AC circuits, use the formula:\[ P = Re(V \cdot I^*) \]In this formula, \( I^* \) stands for the complex conjugate of the current, and \( Re \) designates taking the real part of the product. This calculation effectively takes into account the phase difference, providing the average power consumed over a cycle. This makes it essential in power systems to differentiate between energy that can perform work and energy that simply contributes to system oscillations. Understanding these different forms of power ensures efficient energy management and system design.

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Most popular questions from this chapter

A load consisting of a capacitor in series with a resistor has an impedance of \(50 \Omega\) and a pf of \(0.707\) leading. The load is connected in series with a \(40 \Omega\) resistor across an ac supply and the resulting current is of \(3 \mathrm{~A}\). Determine the supply voltage and overall phase angle. \(\quad\left[249.69 \mathrm{~V}, 25.135^{\circ}\right]\)

An ac circuit has the following voltage and current: \(v=325\) sin \(314 t\), \(i=65 \sin (314 t-1.57) .\) Find (i) frequency, (ii) \(\mathrm{rms}\) value of voltage and current, (iii) impedance, and (iv) power factor.

A choking coil and a pure resistor are connected in series across a supply of \(230 \mathrm{~V}\), \(50 \mathrm{~Hz}\). The voltage drop across the resistor is \(100 \mathrm{~V}\) and that across the chocking coil is \(150 \mathrm{~V}\). Find graphically the voltage drop across the inductance and resistance of the choking coil. Hence, find their values if the current is \(1 \mathrm{~A}\).

Two impedances \(Z_{1}\) and \(Z_{2}\) are connected in series across a \(230 \mathrm{~V}, 50 \mathrm{~Hz}\) ac supply. The total current drawn by the series combination is \(2.3 \mathrm{~A}\). The pf of \(Z_{1}\) is \(0.8\) lagging. The voltage drop across \(Z_{1}\) is twice the voltage drop across \(Z_{2}\) and it is \(90^{\circ}\) out of phase with it. Determine the value of \(Z_{2}\).

A voltage of \(200 \angle 53.13^{\circ} \mathrm{V}\) is applied across two impedances in parallel. The values of the impedances are \((12+j 16) \Omega\) and \((10-j 20) \Omega\). Determine kVA, kVAR and \(\mathrm{kW}\) in each branch and the \(\mathrm{pf}\) of the whole circuit.

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