Kirchhoff s Laws - Lecture notes 1 PDF

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Notes about Kirchhoff s Laws
Teacher: Jones...

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EET 300 || Chapter 1(B) Lesson Notes

Kirchhoff’s Laws

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Kirchhoff’s Laws Kirchhoff’s Voltage Law states that

“The algebraic sum of the voltages around a closed loop must be equal to zero”

or expressed mathematically,



n closed loop

v n   0 “Apply the polarity sign which is attached to

By convention, we:

the entering point of the device.”

“ However, the opposite convention works just as well. JUST BE

+

V

2

-

+

V

3

-

CONSISTENT! Also by convention, we usually draw our loop clockwise. The name of the

+ V 1 -

KVL loop

+ V 4 -

game is consistency. We apply the sign of the device that the loop first touches.

V1  V2  V3  V4  0

V1  V2  V3  V 4

This exemplifies another way of expressing KVL,

“The sum of the voltage rises MUST equal the sum of the voltage drops.”

EET 300 || Chapter 1(B) Lesson Notes

Kirchhoff’s Laws

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“The algebraic sum of Kirchhoff’s Current Law states that,

the currents at a node must equal zero”

By convention, currents entering a node are considered to be NEGATIVE and currents leaving a node are

I2 I1

considered to be POSITIVE.

I3

i1  i2  i 3  0 i1  i2  i3

This demonstrates a more fun way to express KCL:

“The GOES INTA’s must equal the GOES OUTA’s” or “What goes in MUST come out.”

EET 300 || Chapter 1(B) Lesson Notes

Kirchhoff’s Laws

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6k  I1

Example 2: Determine all currents, voltage drops, and the unknown value of R in the circuit shown.

+V13k 

21v

I2

+ -

I4 I3

+

KVL loop 9v

R

-



A clockwise KVL loop is drawn as shown to determine the voltage V1 across the two parallel resistors. By KVL, 

0  21  V1  9

V1  21  9  12v 

Recognize that this voltage is the same across both the 6kΩ and the 3kΩ resistor since they are in parallel. With this voltage, we can determine I2 and I3 with Ohm’s Law.

I2  

12v  2mA 6k 

I3 

12v  4mA 3k 

Next, from KCL, we find the value for I1 to be: 0   I1  I 2  I 3 I1  I2  I 3  2ma  4ma  6mA



It follows by inspection, that: I4  I1  6 mA



Finally, we can calculate the missing resistor value, (R), with Ohm’s Law,

R 

9v  1.5 k  6mA

EET 300 || Chapter 1(B) Lesson Notes

Kirchhoff’s Laws

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I2=2mA



I1=6mA

The circuit as we know it now looks like:

I4=6mA

6k  3k 

21v

+ -

I3 +12v 4mA

+

9v

1.5k 

-



While we have the answers for this example, calculate all the powers involved and determine if the power is being delivered (Sourcing) or absorbed (Sinking). (I use the words interchangeably. )

P21v   21v   6mA   126mW (Sourcing ) P6k  12v   2mA    24mW (Sinking ) P3k  12v   4mA    48mW (Sinking )

P1.5k   9v   6mA    5 4mW (Sinkin g )

0W   126m W  24 mW  48 mW  54 mW

EET 300 || Chapter 1(B) Lesson Notes

Kirchhoff’s Laws

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Equivalent Resistance’s Series Circuits: We need to find a value for Req which will provide the same current I when the same voltage V is applied across it as in the original circuit across the series R’s. I

+ V -

R1 +

R2 V1

+

-

V2

R3 +

V3 -

-

I + Vn -

+ V -

Rn

Req

Start off with a KVL equation:  V  V1  V2  V3      Vn  0

Apply Ohm's Law,  V  IR1  IR2  IR3      IRn  0 Rearrange the equation and factor out I,

I  R1  R2  R3      Rn   V

R1  R2  R3      Rn  V

I

 Re q

So, we note from this that:

The sum of the series resistances is equal to the total resistance of the circuit.

EET 300 || Chapter 1(B) Lesson Notes

Kirchhoff’s Laws

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Parallel Circuits: We need to find a value for Req which will

I

I

provide the same current I when the

+ V R1 -

same voltage V is applied across it as in

I1

R2

I2

R3

I3

the original circuit across the parallel

R’s.



Start out with performing KCL at the top node of the node pair:

0   I  I1  I2   I3      In

I  I1  I 2  I 3      In V V V V    I  R1 R2 R3 Rn 1

I  V 

 R1

Re q 





1

R2



1

R3



1   Rn 

1 V  1 1 1 1 I     R1 R2 R3 Rn

Note that:

The final answer MUST BE SMALLER than the smallest parallel resistor.

Rn

In

+ V -

R eq

EET 300 || Chapter 1(B) Lesson Notes

Kirchhoff’s Laws

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Special case with 2 parallel resistors:

From a bov e 1 1 1   R1 R2 Re q Find the common denomi nato r, R  1 1 1   2    R2 Re q  R2  R1 R2 R1 1   Re q R1R2 R1R2

 R1     R1 

R  R2 1  1 Re q R1R2 Re q

R2

R1

R eq

R1R2 1   R1  R2 R1  R2 R1R2

Voltage Divider Method There is another case where you know a voltage which is across a set of resistors. You can apply a concept known as a Voltage Divider to solve for the voltage across a single resistor with one equation.

RT  R1  R2 E E  IT  RT R1  R2 VR  IT R2 2

Substitute the Equati on for I T  E  ER2 VR    R2  VR  2 2 R1  R2  R1  R2 



IT

+ V1 -

+ VT -

R1 R2

+ V2 -

EET 300 || Chapter 1(B) Lesson Notes

Kirchhoff’s Laws

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Example (Voltage Divider): + V1 R1(5k ) R2 (10k )

+ Calculate the values of all voltages in the figure to the right using the Voltage Divider Rule.

30v -

R3 ) (15k

+ V2 + V3 -

+

V4

-

V1 

30  5k  

5k   10k   15k  150  30  5v

V3 

30 15 k  

30k  450  30  15v

V2 

30 10k  

30k  300  30  10v

V4  

30  10 k   15 k   30 k 

30 25  30

 25v



750 30

EET 300 || Chapter 1(B) Lesson Notes

Kirchhoff’s Laws

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Current Divider Method Like the voltage divider, we can find the

I1

current thru a resistor which is one of several parallel resistors with just one equation called a

R1

IT

current divider.

I2 R1 R2 RT  R1  R2

+

R2 V

-

 RR  V  IT RT  IT  1 2   R1  R2 

I2 

I R V  T T  R2 R2

I2 

IT R1 R1  R2

 R R  1 2   R1  R2   

IT 

R2

Note that the end result was that the current entering the node was multiplied by the opposite resistor and then divided by the sum of the two resistors (not the total resistance).

EET 300 || Chapter 1(B) Lesson Notes

Kirchhoff’s Laws

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I3

Example (Current Divider):

I2

Find the current, I1, by using the Current Divider Method.

IT 10A

6k  3k 

I1

2k 

Note that the parallel equivalent of the 6k and the 3k has to be found first. Why? Because the Current Divider equation is only valid for TWO RESISTORS!

I1  

10A  6k  || 3k 

2k    6k  || 3k   10A  2k  

2k    2k  

20A 4  5A ...