Parallel Circuits Lecture No.5 By – Engr Sajid Hussain Qazi Mehran U.E.T Campus Khairpur.

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Parallel Circuits Lecture No.5 By – Engr Sajid Hussain Qazi Mehran U.E.T Campus Khairpur

Transcript of Parallel Circuits Lecture No.5 By – Engr Sajid Hussain Qazi Mehran U.E.T Campus Khairpur.

Parallel Circuits

Lecture No.5By – Engr Sajid Hussain QaziMehran U.E.T CampusKhairpur

Parallel Circuits

Engr. S.H.Qazi

“Two elements, branches, or networks are in parallel if they have twopoints in common.”

• In Fig. 1, elements 1 and 2 have terminals a and b in common; they are therefore in parallel.

FIGURE-1

• In Fig. 2, all the elements are in parallel because they satisfy the above criterion. Three configurations are provided to demonstrate how the parallel networks can be drawn.

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FIGURE-2

In Fig.3, elements 1 and 2 are in parallel because they have terminals a and b in common. The parallel combination of 1 and 2 is then in series with element 3 due to one common terminal point b.

FIGURE-3

Total Conductance and Resistance• Recall that for series resistors, the total resistance is the sum of the resistor values.

For parallel elements, “The total conductance is the sum of the individual conductance.”

That is, for the parallel network of Fig. 4, we write

nT GGGGG .....321 G-Conductance

FIGURE-4

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Since increasing levels of conductance will establish higher current levels, the more terms appearing in above equation, the higher the input current level.

In other words, as the number of resistors in parallel increases, the input current level will increase for the same applied voltage, the opposite effect of increasing the number of resistors in series.

Substituting resistor values for the network of Fig.4, will result in the network of Fig. 5. Since G=1/R, the total resistance for the network can be determined by direct substitution into Equation.

FIGURE-5Engr. S.H.Qazi

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EXAMPLE.1 Determine the total conductance and resistance for the parallel network of Figure shown.

66.615.0

11

15.020

1

10

1

11

21

TT

T

T

GR

SG

RRG

Solution.

EXAMPLE.2 Determine the effect on the total conductance and resistance of the network of Figure, if another resistor of 10 ohms were added in parallel with the other elements.

4

25.0

11

25.010

115.0

T

TT

T

T

R

GR

SG

G

Solution.

Note, as mentioned above, that adding additional terms increases the conductance level and decreases the resistance level.

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EXAMPLE .3 Determine the total resistance for the network shown below.

For same value of resistors, the equation for finding Rt, becomes more easier to apply. For N equals resistors in parallel, equation becomes..

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EXAMPLE .4 Determine the total resistance for the networks shown below.

Answer-4 ohms

Answer-0.5 ohms

When number of resistors in circuit are two, then total resistance can be found by using below formula.

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Example.5. (a) Determine total resistance of circuit shown in fig.1. (b) What is the effect on total resistance of fig.1 if an additional resistor of same value is added in parallel as shown in fig.2. (c) What is the effect on total resistance of fig.1 if a large resistance is added in parallel as shown in fig.3. (d) What is the effect on total resistance of fig.1 if a small resistance is added in parallel as shown in fig.4.

Fig.1

Fig.4

Fig.3

Fig.2

Exercise Problem

(15 ohm)

(0.099 ohm)(10 ohm)

(14.77 ohm)

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Current Divider Rule• As the name suggests, the current divider rule (CDR) will determine how the current entering a set of parallel branches will split between the elements.

“For two parallel elements of equal value, the current will divide equally.”

“For parallel elements with different values, the smaller the resistance, the greater the share of input current.”

“For parallel elements of different values, the current will split with a ratio equal to the inverse of their resistor values.”

• For example, if one of two parallel resistors is twice the other, then the current through the larger resistor will be half the other.

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In Fig. 6, since I1 is 1 mA and R1 is six times R3, the current through R3 must be 6 mA . For R2 the current must be 2 mA since R1 is twice R2. The total current must then be the sum of I1, I2, and I3.

FIGURE-6

For networks in which only the resistor values are given along with the input current, the current divider rule should be applied to determine the various branch currents. It can be derived using the network of Fig. 7.

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Input Current I equals

TR

VI

Substituting V=IxRx in above equation

T

xx

R

RII

This is the general form for the current divider rule. In words, the current through any parallel branch is equal to

“The product of the total resistance of the parallel branches and the input current divided by the resistance of the branch through which the

current is to be determined.”

For the current I1,

IR

RI T

11 I

R

RI T

22

For the current I2,

And so on….

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For the particular case of two parallel resistors, as shown in Fig.7.

FIGURE-7

IRR

RI *

21

21

For I2,

IRR

RI *

21

12

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Example 6. Determine current I2 for network shown.

Example 7. Determine current I1 for circuit shown.

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Example 8. Determine current I1, I2 and I3 for circuit shown. Also verify KCL.

Using CDR

Applying KCL

For verification, using CDR

Total current entering the branch must equal to leaving..

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Example 9. Determine resistance R1 for circuit shown.

Practice Example

(R2=2 ohm)

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PRACTICE TESTQ.1. Two branches will be in parallel if they have ______ terminals in common.

TWO

Q.2. Opposite of resistance is_______ and its unit is __________.

Conductance/ Siemen

Q.3. If number of parallel resistors increases the value of _______________ increases for the same applied __________.

Input Current/Voltage

Q.4. In parallel circuits relation of current and resistance is ___________.

Inverse