Parallel RF RLC Circuits

Problem Solving

Current

• The easiest way of evaluating parallel RLC circuits is by the total current method. We calculate the current in each branch and then determine their vector sum. We begin by finding the opposition in each branch. (Refer to Fig. 1A)

L1150mH

1

R1220k

C10.05uF

V1

8 V 2.5kHz 0Deg

0

Figure 1A. Circuit

2355

10150105.228.6

233 HXHzXXX

fLX L

1274785.0

101

1005.0105.228.6

1

2

1

3

63

X

FXHzXXX

fCXC

• The current in each branch must be calculated. Keep in mind that we must use both magnitude and phase.

A

V

X

EI

L

TL

0

0

0

9000340.0

902355

08

A

V

X

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C

TC

0

0

0

9000628.0

901274

08

A

V

R

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0

0

0

000364.0

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• The total current equals the vector sum of all branch currents. (Refer to Fig. 1B)

A

AA

III LCXT

0

00

9000288.0

9000340.09000628.0

IR= 0.00364 A

IL=0.00340 A

IXT=0.00288 A

IC=0.00628 A

IT

Figure 1B. Vector Sum of Circuit Currents: Parallel RLC Circuits

A

A

A

A

AIII XTRT

00464.0

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00288.000364.0 22

22

03.38

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00288.0

tan

A

A

I

IA

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XT

ampereITherefore

A

AII

I

I

H

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T

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03.3800464.0,

00464.06198.0

00288.0

sin

sin

• This circuit is capacitive because the current leads the voltage. Note that the smallest reactance in a parallel circuit dominates, since it has the greatest current flow.

Impedance

• Once the value of total current is known, impedance can be determined. As always, the impedance of a circuit equals the total voltage divided by the total current. The total current method continues to be our basic method of evaluating AC circuits.

0

0

0

3.381724

3.3800464.0

08

A

V

I

EZ

T

T

Frequency Response

• The best way to consider the circuit’s response to frequencies is to change the frequency from 2500 hertz to, say, 1070 hertz and repeat all calculations. If no other circuit values are changed, we will see what effect the frequency change has.

1008

101501007.128.6

233 HXHzXXX

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1

2

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A

V

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0

0

0

9000794.0

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08

A

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0

0

0

9000269.0

902976

08

A

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0

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A

AA

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0

00

9000525.0

9000269.09000794.0

• Fig. 2 shows the capacitor current at +900, the inductor current at -900, and the vector difference also at - 900, since the inductor current is larger. Remember that the larger current dominates in a parallel circuit. The total current is the vector sum of the resistor current at 00 and the net inductor current.

IR= 0.00364 A

IL=0.00794 A

IXT=0.00525 A

IC=0.00269 A

Figure 2. Vector Sum of Currents at Second Frequency

A

A

A

A

III XTRT

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00525.000364.0 22

22

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00525.0

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A

A

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XT

ampereITherefore

AA

II

I

I

H

O

T

XTT

T

XT

03.5500639.0,

00639.08821.0

00525.0sin

sin

• In response to the frequency change, capacitive reactance has become larger and its current smaller. Inductive reactance has become smaller and its current larger. Thus, the circuit is now inductive. This result is also evident from the fact that the current is lagging the voltage.

• Another change occurs in the circuit impedance.

0

0

0

3.551252

3.5500639.0

08

A

V

I

EZ

T

T

Power

• The first step in calculating the value of the true power for this circuit is to determine the power factor. It is equal to the cosine of the phase angle.

Power factor = cos 55.30 =0.5693

• Thus 56.93% of the energy supplied to the circuit is dissipated in some form, such as heat.

The true power can be determined in two ways.

W

XAXA

RIP

W

XAXV

EIP

0291.0

)2200()00364.0()00364.0(

0291.0

)5693.0()00639.0()8(

cos

2

• The total current is not used in the second formula because that current does not pass through the resistor. Only the resistor branch current is used. However, total current is used in the formula

cosEIP