Rlc And Glc Circuit Analysis Quiz

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

10 + j3.00

10 + j29.2

10 − j97

10 + j348

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.

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.

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

150 + j100

150 − j200

100 − j200

150 − j100

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.

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.

3. 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?

1.18 A

1.13 A

0.886 A

0.846 A

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.

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.

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

2.00 A

2.40 A

1.33 A

0.800 A

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.

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.

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

0 + j467

25 + j100

0 − j467

25 − j100

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.

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.

6. 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?

0 − j0.02

0 − j0.07

0 + j0.02

−0.05 + j0.03

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.

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.

7. A vector pointing southeast in the GB plane would indicate

Pure conductance with zero susceptance.

Conductance and inductive susceptance.

Conductance and capacitive susceptance.

Pure susceptance with zero conductance.

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.

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.

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

Always points straight up.

Always points straight down.

Always points straight toward the right.

None of the above is correct.

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.

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.

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

0.0044 + j 0.024

0.035 − j0.011

−0.011 + j0.035

0.0044 + j0.046

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.

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.

10. 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?

100 + j0.00354

0.010 + j0.00354

100 − j0.0144

0.010 + j0.0144

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.

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.

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

0 + j550

0 − j50

250 − j300

−300 + j250

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.

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.

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

50 + j5.0

0.495 − j4.95

50 − j5.0

0.495 + j4.95

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.

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.

13. What is the current through the resistance in the preceding example?

2.00 A

2.40 A

1.33 A

0.800 A

The current through the resistance in the preceding example is 2.00 A.

Explanation

The current through the resistance in the preceding example is 2.00 A.

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

0 + j0.25

0 + j4.00

0 − j0.25

0 − j4.00

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.

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.

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

1.0 + j0.0

1.0 + j1.5

1.0 − j1.5

1.0 − j2.0

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.

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.

16. What is the voltage across the reactance in the preceding example?

78.0 V

55.1 V

99.4 V

74.4 V

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

Explanation

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

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

51.0 − j14.9

51.0 + j14.9

46.2 − j14.9

46.2 + j14.9

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.

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.