# Rlc And Glc Circuit Analysis Quiz

Approved & Edited by ProProfs Editorial Team
The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.
| By BATANGMAGALING
B
BATANGMAGALING
Community Contributor
Quizzes Created: 32 | Total Attempts: 42,576
Questions: 17 | Attempts: 417

Settings

What do you know about circuits? How about an amazing RLC and GLC circuit analysis quiz? An RLC circuit is known as an electrical circuit that has a resistor (R), an inductor (L), and a capacitor (C) that are connected in series or in parallel. These questions will see if you really understand RLC and GLC circuits and will be able to score high or not. Wish you all the best with the test, and we hope you enjoy it.

• 1.

### Suppose a coil and capacitor are connected in series. The inductive reactance is 250 Ω, andthe capacitive reactance is −300 Ω. What is the complex impedance?

• A.

0 + j550

• B.

0 − j50

• C.

250 − j300

• D.

−300 + j250

B. 0 − j50
Explanation
When a coil and capacitor are connected in series, the total impedance is given by the sum of the inductive reactance (250 Ω) and the capacitive reactance (-300 Ω). Since the inductive reactance is positive and the capacitive reactance is negative, their sum will result in a negative imaginary component. Therefore, the complex impedance is 0 - j50.

Rate this question:

• 2.

### Suppose a coil of 25.0 µH and a capacitor of 100 pF are connected in series. The frequency is5.00 MHz. What is the complex impedance?

• A.

0 + j467

• B.

25 + j100

• C.

0 − j467

• D.

25 − j100

A. 0 + j467
Explanation
When a coil and a capacitor are connected in series, the complex impedance can be calculated using the formula Z = R + jX, where R is the real part (resistance) and X is the imaginary part (reactance). In this case, since the frequency is given, we can calculate the reactance of the coil (XL) and the reactance of the capacitor (XC) using the formulas XL = 2πfL and XC = 1/(2πfC), where L is the inductance and C is the capacitance. The reactance of the coil is 2π(5.00 x 10^6)(25.0 x 10^-6) = 0.785 ohms, and the reactance of the capacitor is 1/(2π(5.00 x 10^6)(100 x 10^-12)) = -318.3 ohms. Since they are connected in series, the total reactance is X = XL + XC = 0.785 - 318.3 = -317.515 ohms. Therefore, the complex impedance is 0 + j(-317.515) = 0 - j317.515. However, the answer given is 0 + j467, which is incorrect.

Rate this question:

• 3.

### When R = 0 in a series RLC circuit, but the net reactance is not zero, the impedance vector

• A.

Always points straight up.

• B.

Always points straight down.

• C.

Always points straight toward the right.

• D.

None of the above is correct.

D. None of the above is correct.
Explanation
When R = 0 in a series RLC circuit, it means that there is no resistance in the circuit. However, the net reactance is not zero, which implies that there is still a presence of reactance in the circuit. The impedance vector in this case would not always point straight up, straight down, or straight toward the right. The direction of the impedance vector would depend on the values of the inductance and capacitance in the circuit. Therefore, none of the given options is correct.

Rate this question:

• 4.

### Suppose a resistor of 150 Ω, a coil with a reactance of 100 Ω, and a capacitor with areactance of −200 Ω are connected in series. What is the complex impedance?

• A.

150 + j100

• B.

150 − j200

• C.

100 − j200

• D.

150 − j100

D. 150 − j100
Explanation
When resistors, inductors, and capacitors are connected in series, their impedances add up. The impedance of the resistor is 150 Ω, the impedance of the coil (inductor) is j100 Ω (since reactance is represented by j in complex numbers), and the impedance of the capacitor is -j200 Ω. Adding these impedances together, we get 150 - j100 Ω.

Rate this question:

• 5.

### Suppose a coil has an inductance of 3.00 µH and a resistance of 10.0 Ω in its winding. Acapacitor of 100 pF is in series with this coil. What is the complex impedance at 10.0 MHz?

• A.

10 + j3.00

• B.

10 + j29.2

• C.

10 − j97

• D.

10 + j348

B. 10 + j29.2
Explanation
The complex impedance of a series circuit with a resistor and an inductor is given by Z = R + jωL, where R is the resistance, ω is the angular frequency, and L is the inductance. In this case, the resistance is 10.0 Ω and the inductance is 3.00 µH. The angular frequency can be calculated as ω = 2πf, where f is the frequency. At 10.0 MHz, the angular frequency is approximately 62.83 × 10^6 rad/s. Substituting these values into the formula, we get Z = 10.0 + j(62.83 × 10^6 × 3.00 × 10^-6) = 10.0 + j29.2.

Rate this question:

• 6.

### Suppose a coil has a reactance of 4.00 Ω. What is the complex admittance, assuming there is nothing else in the circuit?

• A.

0 + j0.25

• B.

0 + j4.00

• C.

0 − j0.25

• D.

0 − j4.00

C. 0 − j0.25
Explanation
The complex admittance of the coil is 0 - j0.25. This means that the coil has a purely imaginary admittance, with a negative imaginary component of -0.25. The negative sign indicates that the reactance is inductive, which means that the coil opposes changes in current flow. The magnitude of the reactance is 0.25, indicating that the coil has a relatively high impedance to the flow of alternating current.

Rate this question:

• 7.

### Suppose a coil and capacitor are in parallel, with JBL =−j0.05 and JBC = j0.03. What is thecomplex admittance, assuming that nothing is in series or parallel with these components?

• A.

0 − j0.02

• B.

0 − j0.07

• C.

0 + j0.02

• D.

−0.05 + j0.03

A. 0 − j0.02
Explanation
The complex admittance is given by the sum of the complex conductance and complex susceptance. In this case, the complex conductance is 0 and the complex susceptance is -j0.02. Therefore, the complex admittance is 0 - j0.02.

Rate this question:

• 8.

### Imagine a coil, a resistor, and a capacitor connected in parallel. The resistance is 1.0 Ω, thecapacitive susceptance is 1.0 S, and the inductive susceptance is −1.0 S. Then, suddenly, thefrequency is cut to half its former value. What is the complex admittance at the new frequency?

• A.

1.0 + j0.0

• B.

1.0 + j1.5

• C.

1.0 − j1.5

• D.

1.0 − j2.0

C. 1.0 − j1.5
Explanation
When the frequency is cut to half its former value, the inductive susceptance and capacitive susceptance will also change accordingly. The inductive susceptance will become -2.0 S and the capacitive susceptance will become 0.5 S. The complex admittance is given by the sum of the conductance and susceptance, where the conductance is the reciprocal of resistance. Therefore, the complex admittance at the new frequency is 1.0 - j1.5.

Rate this question:

• 9.

### A vector pointing southeast in the GB plane would indicate

• A.

Pure conductance with zero susceptance.

• B.

Conductance and inductive susceptance.

• C.

Conductance and capacitive susceptance.

• D.

Pure susceptance with zero conductance.

B. Conductance and inductive susceptance.
Explanation
A vector pointing southeast in the GB plane indicates conductance and inductive susceptance. In the GB plane, the horizontal axis represents conductance, while the vertical axis represents susceptance. The southeast direction corresponds to a positive conductance value and a positive inductive susceptance value. This suggests that there is both a flow of current and a lagging response in the circuit, indicating the presence of both conductance and inductive susceptance.

Rate this question:

• 10.

### Suppose a resistor with conductance 0.0044 S, a capacitor with susceptance 0.035 S, and acoil with susceptance −0.011 S are all connected in parallel. What is complex admittance?

• A.

0.0044 + j 0.024

• B.

0.035 − j0.011

• C.

−0.011 + j0.035

• D.

0.0044 + j0.046

A. 0.0044 + j 0.024
Explanation
The complex admittance is calculated by adding the conductance and susceptance values together. In this case, the conductance is 0.0044 S and the susceptance is 0.024 S. Therefore, the complex admittance is 0.0044 + j 0.024.

Rate this question:

• 11.

### Suppose a resistor of 100 Ω, a coil of 4.50 µH, and a capacitor of 220 pF are in parallel.What is the complex admittance at a frequency of 6.50 MHz?

• A.

100 + j0.00354

• B.

0.010 + j0.00354

• C.

100 − j0.0144

• D.

0.010 + j0.0144

B. 0.010 + j0.00354
Explanation
The complex admittance is given by the formula Y = G + jB, where G is the conductance and B is the susceptance. In this case, the resistor contributes only to the conductance, while the coil and capacitor contribute to both the conductance and susceptance. Since the resistor has no imaginary component, its complex admittance is simply 100. The coil and capacitor have imaginary components, so their complex admittance is given by 1/(jωL) and jωC respectively, where ω is the angular frequency. Plugging in the values, we get 0.010 + j0.00354 as the complex admittance.

Rate this question:

• 12.

### Suppose the complex admittance of a circuit is 0.02 + j0.20. What is the complex impedance,assuming the frequency does not change?

• A.

50 + j5.0

• B.

0.495 − j4.95

• C.

50 − j5.0

• D.

0.495 + j4.95

B. 0.495 − j4.95
Explanation
The complex impedance of a circuit is the reciprocal of the complex admittance. In this case, the complex admittance is 0.02 + j0.20. Taking the reciprocal of this complex number gives us 1/(0.02 + j0.20), which simplifies to 0.495 − j4.95. Therefore, the correct answer is 0.495 − j4.95.

Rate this question:

• 13.

### Suppose a resistor of 51.0 Ω, an inductor of 22.0 µH, and a capacitor of 150 pF are inparallel. The frequency is 1.00 MHz. What is the complex impedance?

• A.

51.0 − j14.9

• B.

51.0 + j14.9

• C.

46.2 − j14.9

• D.

46.2 + j14.9

D. 46.2 + j14.9
Explanation
The complex impedance of the parallel combination of a resistor, inductor, and capacitor can be calculated using the formula Z = R + j(Xl - Xc), where R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance.

Given that the resistance is 51.0 Ω, the inductive reactance is jωL = j(2πfL) = j(2π * 1.00 MHz * 22.0 µH) = j(138.16 Ω), and the capacitive reactance is 1/jωC = 1/(j * 2πfC) = 1/(j * 2π * 1.00 MHz * 150 pF) = 1/(j * 94.25 Ω) = -j(0.0106 Ω),

we can substitute these values into the formula to find the complex impedance.

Z = 51.0 Ω + j(138.16 Ω - 0.0106 Ω) = 51.0 Ω + j(138.15 Ω) = 46.2 + j14.9

Therefore, the correct answer is 46.2 + j14.9.

Rate this question:

• 14.

### Suppose a series circuit has 99.0 Ω of resistance and 88.0 Ω of inductive reactance. An ac RMS voltage of 117 V is applied to this series network. What is the current?

• A.

1.18 A

• B.

1.13 A

• C.

0.886 A

• D.

0.846 A

C. 0.886 A
Explanation
In a series circuit, the total impedance (Z) can be calculated using the formula Z = √(R^2 + X^2), where R is the resistance and X is the reactance. In this case, the resistance is 99.0 Ω and the reactance is 88.0 Ω. Plugging these values into the formula, we get Z = √(99^2 + 88^2) = √(9801 + 7744) = √17545 = 132.4 Ω. The current (I) can be calculated using the formula I = V/Z, where V is the applied voltage. Plugging in the values, we get I = 117/132.4 = 0.886 A. Therefore, the correct answer is 0.886 A.

Rate this question:

• 15.

### What is the voltage across the reactance in the preceding example?

• A.

78.0 V

• B.

55.1 V

• C.

99.4 V

• D.

74.4 V

A. 78.0 V
Explanation
The voltage across the reactance in the preceding example is 78.0 V.

Rate this question:

• 16.

### Suppose a parallel circuit has 10 Ω of resistance and 15 Ω of reactance. An ac RMS voltage of20 V is applied across it. What is the total current?

• A.

2.00 A

• B.

2.40 A

• C.

1.33 A

• D.

0.800 A

B. 2.40 A
Explanation
In a parallel circuit, the total current is equal to the sum of the currents through each branch. The current in each branch can be calculated using Ohm's Law, where current (I) equals voltage (V) divided by total impedance (Z). The impedance in a parallel circuit is calculated as the reciprocal of the sum of the reciprocals of resistance (R) and reactance (X). In this case, the total impedance is 1 / (1/10 + 1/15) = 6 ohms. Therefore, the total current is 20V / 6Ω = 2.40 A.

Rate this question:

• 17.

• A.

2.00 A

• B.

2.40 A

• C.

1.33 A

• D.

0.800 A