AC Theory Level 2, Lesson 6: Understanding and Analyzing Parallel RLC Circuits in Electrical Engineering

A parallel RLC circuit must contain at least three branches.

True

For example

Br1 `- L

Br2 - C

Br3 - R

6Q1

In parallel RLC circuits, the ? is the same and is ? phase across all branches and is used as the main reference

for calculations.

voltage / in

Note: This question references only the phase relationship with ET, ER, EL, and EC.

6Q2

Considering the phase relationship between current and voltage in a parallel RLC circuit, the resistive current is ? .

0° in-phase

6Q3

Considering the phase relationship between current and voltage in a parallel RLC circuit, the inductive current is ? .

90 degrees lagging

6Q4

Considering the phase relationship between current and voltage in a parallel RLC circuit, the capacitive current is ? .

90 degrees leading

6Q5

True or False

In a parallel circuit that contains both XL and XC, the reactive currents add.

False

Note: Current in an inductive branch is 180° out-of-phase with current in capacitive branch; therefore, the currents will subtract.

6Q6

Given a simple RLC circuit

E = 150 V

F = 60 Hz

X_L_ = 60 ohms

X_C_ = 30 ohms

R = 50 ohms

Find current for the resistive branch

I_R_ = E / R

I_R_ = 150 / 50

I_R_ = 3 A

6Q7

Given a simple RLC circuit

E = 150 V

F = 60 Hz

X_L_ = 60 ohms

X_C_ = 30 ohms

R = 50 ohms

Find current for the inductive branch

I_L_ = E / X_L_

I_L_ = 150 / 60

I_L_ = 2.5 A

6Q8

Given a simple RLC circuit

E = 150 V

F = 60 Hz

X_L_ = 60 ohms

X_C_ = 30 ohms

R = 50 ohms

Find current for the capacitive branch

I_C_ = E / X_C_

I_C_ = 150 / 30

I_C_ = 5 A

6Q9

Given a simple RLC circuit

E = 150 V

F = 60 Hz

X_L_ = 60 ohms

X_C_ = 30 ohms

R = 50 ohms

What is the phase relationship b/w I_L_ and I_C_ shown

180 degrees b/c

IL lags E by 90°, and IC leads E by 90°; therefore, IL and IC are 180° out-of-phase with respect to each other.

6Q10

Given a simple rlc crkt

E = 120 V

F = 60 Hz

I_L_ = 14 A

I_C_ = 6 A

I_R_ = 12 A

What is the true power of the circuit shown?

P_true_ = e * I_R_

= 120 * 12

= 1,440 W

6Q11

Given a simple rlc crkt

E = 120 V

F = 60 Hz

I_L_ = 14 A

I_C_ = 6 A

I_R_ = 12 A

Solve for the net reactive current (I_X_).

I_X_ = I_L_ - I_C_

=14 - 6

= 8 A

6Q12

Given a simple rlc crkt

E = 120 V

F = 60 Hz

I_L_ = 14 A

I_C_ = 6 A

I_R_ = 12 A

Solve for the reactive power (vars).

P_reactive_ = E * ( I_L_ - I_C_)

=120 * (14-6)

=960 var

6Q13

Given a simple rlc crkt

E = 120 V

F = 60 Hz

I_L_ = 14 A

I_C_ = 6 A

I_R_ = 12 A

Solve for the apparent power (VA).

1731 VA

6Q14

Given a -45 degree power triangle.

What side is adjacent?

Opposite?

Hypotenuse?

Adj = 1440W

Opp = 960var

Hyp = 1731VA

6Q15

True or False

The formula I_T_ = sqrt ( I_R_^2 = (I_L_ -I_C_)^2 ) can be used to find total current in a parallel RLC circuit using branch currents.

True

6Q16

Use the branch currents to determine the total current of the circuit shown. The values calculated for this question may be used for additional questions. (Round the FINAL answer to one decimal place.)

14.4 A

Given a simple rlc crkt

E = 120 V

F = 60 Hz

I_L_ = 14 A

I_C_ = 6 A

R_ = 12 A

What is the apparent power?

VA = E_T_ * I_T_

VA = 120 * 14.4

VA = 1,728 VA

6Q18

True or False

Different methods of calculating circuit values may produce answers that are slightly different due to rounding of the numbers used as components of any particular formula as it is executed.

True

6Q19

The impedance of a parallel RLC circuit is 40 ohms, and the total current is 3 amperes. What is the applied voltage?

E_app_ = I_T_ * Z\

E_app_ = 3 * 40

E_app_ = 120 V

6Q20

Given a simple rlc crkt

E = 120 V

f = 60 Hz

X_L_ = 30 ohms

X_C_ = 40 ohms

R = 20 ohms

Find

I_R_

I_L_

I_C_

I_T_

angle theta

Z

PF

E_R_ = E = E_L_ = E_C_ = 120

I_R_ = E_R_ / R

I_R_ = 120 / 20

I_R_ = 6

I_L_ = E_L_ / X_L_

I_L_ = 120 / 30

I_L_ = 4

I_C_ = E_C_ / X_C_

I_C_ = 120 / 40

I_C_ = 3

I_T_ = sqrt( I_R_^2 = (I_L_ -I_C_)^2 )

I_T_ = sqrt( 6^2 = (4 -3)^2 )

I_T_ = 6.08 A

Z = 19.73Ω

angle theta = atan<(I_L_ -I_C_)/ I_R_ >

angle theta = atan<(4-3) / 6>

angle theta = 9.5 degrees

PF = cos(angle theta) * 100

PF = cos(9.46) *100

PF = 99%

6Q21

Place the labels in the appropriate locations so that the vector diagram properly indicates the values for IR, IC, IL, IT for the parallel RLC circuit shown. (Be sure to place variables to the left and values to the right of the equals sign.)

Top to bottom

IC = 3A

IR = 6A

IT = 6.08 A

IL = 4A

Resonance occurs at the particular frequency that would cause X_L_ to be equal to X_C_ in a circuit.

True

6Q23

In a parallel circuit, inductive current and capacitive current are ? out-of-phase.

180 degrees

6Q24

True or False

In a theoretical parallel circuit having only pure inductance and capacitance operating at resonant frequency, some amount of power is consumed.

False Note:

No power is consumed in a circuit having only inductance and capacitance because the currents are 180° out-of-phase and their values are equal at resonant frequency.

6Q25

True or False

A parallel circuit having resistance, inductance, and capacitance is sometimes known as a tank circuit.

True

6Q26

Given a simple rlc crkt

If the frequency was increased towards resonace and...

E = 120 V

F = 60 Hz

X_L_ = 14 A

X_C_ = 6 A

R_ = 12 A

R would ...

X_L_ would...?

X_C_ would...?

Z would... ?

I_T_ would ... ?

The power factor would ... ?

R would stay the same

X_L_ would increase

X_C_ would decrease

Z would increase

I_T_ would decrease

PF would increase

6Q27

Using the formula f_r_ = 1 / 2 / pi / sqrt (LC) , where f_r_ is frequency at resonance, L is inductance in henries, and C is capacitance in farads, calculate the resonant frequency for a circuit supplied by a 200-volt AC source, a 0.500-microfarad capacitor, and a 0.05-henry inductor in parallel. The values calculated for this question may be used for additional questions. (Round the FINAL answer to the nearest whole number.)

f_r_ = 1 / 2 / sqrt ( .05 * .0000005)

=1,007 Hz

6Q28

Given

E = 200 V

f = 1007Hz

L = 0.05 H

C = 500 nF

Solve for ....

X_L_

X_C_

I_C_

I_L_

I_T_

X_L_ = 2 pi f L

X_L_ = 2 pi 1007 * .05

X_L_ = 316

X_C_ = 1 / 2 / pi / f / C

X_C_ = 1 / 2/ pi / 1007 / .0000005

X_C_ = 316

I_C_ = E / X_C_

I_C_ = 200 / 316

I_C_ = 630 mA

I_L_ = E / X_L_

I_L_ = 200 / 316

I_L_ = 630 mA

I_T_ = 0 A

6Q29