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Divider Circuits and Kirchhoff’s Laws - Review Sheet | PHYS 3036, Study notes of Physics

Material Type: Notes; Professor: Zhou; Class: ELEMENTARY ELECTRONICS LAB; Subject: Physics; University: Montana Tech of the University of Montana; Term: Unknown 1989;

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Chapter 6
DIVIDER CIRCUITS AND
KIRCHHOFF’S LAWS
Contents
6.1 Voltage divider circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
6.2 Kirchhoff’s Voltage Law (KVL) . . . . . . . . . . . . . . . . . . . . . . . 173
6.3 Current divider circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
6.4 Kirchhoff’s Current Law (KCL) . . . . . . . . . . . . . . . . . . . . . . . 187
6.5 Contributors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
6.1 Voltage divider circuits
Let’s analyze a simple series circuit, determining the voltage drops across individual resistors:
+
-
R1
R2
R3
5 k
7.5 k
10 k45 V
165
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a

Partial preview of the text

Download Divider Circuits and Kirchhoff’s Laws - Review Sheet | PHYS 3036 and more Study notes Physics in PDF only on Docsity!

Chapter 6

DIVIDER CIRCUITS AND

KIRCHHOFF’S LAWS

Contents

6.1 Voltage divider circuits............................ 165 6.2 Kirchhoff ’s Voltage Law (KVL)....................... 173 6.3 Current divider circuits............................ 184 6.4 Kirchhoff ’s Current Law (KCL)....................... 187 6.5 Contributors.................................. 189

6.1 Voltage divider circuits

Let’s analyze a simple series circuit, determining the voltage drops across individual resistors:

R 1

R 2

R 3

5 kΩ

7.5 kΩ

45 V 10 kΩ

165

166 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS

E

I

R

Volts Amps Ohms

R 1 R 2 R 3 Total

5k 10k 7.5k

From the given values of individual resistances, we can determine a total circuit resistance, knowing that resistances add in series:

E

I

R

Volts Amps Ohms

R 1 R 2 R 3 Total

5k 10k 7.5k 22.5k

From here, we can use Ohm’s Law (I=E/R) to determine the total current, which we know will be the same as each resistor current, currents being equal in all parts of a series circuit:

E

I

R

Volts Amps Ohms

R 1 R 2 R 3 Total

5k 10k 7.5k

22.5k

2m 2m 2m 2m

Now, knowing that the circuit current is 2 mA, we can use Ohm’s Law (E=IR) to calculate voltage across each resistor:

E

I

R

Volts Amps Ohms

R 1 R 2 R 3 Total

5k 10k 7.5k

22.5k

2m 2m 2m 2m

It should be apparent that the voltage drop across each resistor is proportional to its resistance, given that the current is the same through all resistors. Notice how the voltage across R 2 is double that of the voltage across R 1 , just as the resistance of R 2 is double that of R 1. If we were to change the total voltage, we would find this proportionality of voltage drops remains constant:

E

I

R

Volts Amps Ohms

R 1 R 2 R 3 Total

5k 10k 7.5k 22.5k

8m 8m 8m 8m

168 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS

drop to total supply voltage in a voltage divider circuit. This is known as the voltage divider formula, and it is a short-cut method for determining voltage drop in a series circuit without going through the current calculation(s) of Ohm’s Law. Using this formula, we can re-analyze the example circuit’s voltage drops in fewer steps:

R 1

R 2

R 3

5 kΩ

7.5 kΩ

45 V 10 kΩ

ER1 = 5 kΩ 22.5 kΩ

45 V = 10 V

ER2 =45 V

22.5 kΩ

10 kΩ (^) =20 V

ER3 =45 V

22.5 kΩ

7.5 kΩ (^) =15 V

Voltage dividers find wide application in electric meter circuits, where specific combinations of se- ries resistors are used to ”divide” a voltage into precise proportions as part of a voltage measurement device.

Input

voltage Divided

voltage

R 1

R 2

One device frequently used as a voltage-dividing component is the potentiometer, which is a resistor with a movable element positioned by a manual knob or lever. The movable element, typically called a wiper, makes contact with a resistive strip of material (commonly called the slidewire if made of resistive metal wire) at any point selected by the manual control:

6.1. VOLTAGE DIVIDER CIRCUITS 169

1

wiper contact

Potentiometer

The wiper contact is the left-facing arrow symbol drawn in the middle of the vertical resistor element. As it is moved up, it contacts the resistive strip closer to terminal 1 and further away from terminal 2, lowering resistance to terminal 1 and raising resistance to terminal 2. As it is moved down, the opposite effect results. The resistance as measured between terminals 1 and 2 is constant for any wiper position.

1

less resistance

more resistance

less resistance

more resistance

Shown here are internal illustrations of two potentiometer types, rotary and linear:

Resistive strip

Wiper

Terminals

Rotary potentiometer construction

6.1. VOLTAGE DIVIDER CIRCUITS 171

If a constant voltage is applied between the outer terminals (across the length of the slidewire), the wiper position will tap off a fraction of the applied voltage, measurable between the wiper contact and either of the other two terminals. The fractional value depends entirely on the physical position of the wiper:

more voltage less voltage

Using a potentiometer as a variable voltage divider

Just like the fixed voltage divider, the potentiometer’s voltage division ratio is strictly a function of resistance and not of the magnitude of applied voltage. In other words, if the potentiometer knob or lever is moved to the 50 percent (exact center) position, the voltage dropped between wiper and either outside terminal would be exactly 1/2 of the applied voltage, no matter what that voltage happens to be, or what the end-to-end resistance of the potentiometer is. In other words, a potentiometer functions as a variable voltage divider where the voltage division ratio is set by wiper position. This application of the potentiometer is a very useful means of obtaining a variable voltage from a fixed-voltage source such as a battery. If a circuit you’re building requires a certain amount of voltage that is less than the value of an available battery’s voltage, you may connect the outer terminals of a potentiometer across that battery and ”dial up” whatever voltage you need between the potentiometer wiper and one of the outer terminals for use in your circuit:

172 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS

Circuit requiring less voltage than what the battery provides

V

Adjust potentiometer to obtain desired voltage Battery

When used in this manner, the name potentiometer makes perfect sense: they meter (control) the potential (voltage) applied across them by creating a variable voltage-divider ratio. This use of the three-terminal potentiometer as a variable voltage divider is very popular in circuit design.

Shown here are several small potentiometers of the kind commonly used in consumer electronic equipment and by hobbyists and students in constructing circuits:

The smaller units on the very left and very right are designed to plug into a solderless breadboard or be soldered into a printed circuit board. The middle units are designed to be mounted on a flat panel with wires soldered to each of the three terminals.

Here are three more potentiometers, more specialized than the set just shown:

174 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS

implied, for positive readings in digital meter displays. However, for this lesson the polarity of the voltage reading is very important and so I will show positive numbers explicitly:

E2-1 = +45 V

When a voltage is specified with a double subscript (the characters ”2-1” in the notation ”E 2 − 1 ”), it means the voltage at the first point (2) as measured in reference to the second point (1). A voltage specified as ”Ecg ” would mean the voltage as indicated by a digital meter with the red test lead on point ”c” and the black test lead on point ”g”: the voltage at ”c” in reference to ”g”.

A COM

V

V A

OFF A

d c

Ecd

The meaning of

Black Red

If we were to take that same voltmeter and measure the voltage drop across each resistor, stepping around the circuit in a clockwise direction with the red test lead of our meter on the point ahead and the black test lead on the point behind, we would obtain the following readings:

E3-2 = -10 V

E4-3 = -20 V

E1-4 = -15 V

6.2. KIRCHHOFF’S VOLTAGE LAW (KVL) 175

R 1

R 2

R 3

5 kΩ 10 kΩ 7.5 k Ω

A^ V COM^ Ω 45 V

A^ V COM^ Ω

A^ V COM^ Ω

A COM^ V^ Ω

E2-

E3-

E4-

E1-

We should already be familiar with the general principle for series circuits stating that individual voltage drops add up to the total applied voltage, but measuring voltage drops in this manner and paying attention to the polarity (mathematical sign) of the readings reveals another facet of this principle: that the voltages measured as such all add up to zero:

-10 V

-20 V

-15 V

+45 V

0 V

voltage from point 2 to point 1 voltage from point to point voltage from point to point voltage from point to point

E2-1 =

E3-2 =

E4-3 =

E1-4 =

This principle is known as Kirchhoff ’s Voltage Law (discovered in 1847 by Gustav R. Kirchhoff, a German physicist), and it can be stated as such:

”The algebraic sum of all voltages in a loop must equal zero”

By algebraic, I mean accounting for signs (polarities) as well as magnitudes. By loop, I mean any path traced from one point in a circuit around to other points in that circuit, and finally back to the initial point. In the above example the loop was formed by following points in this order: 1-2-3-4-1. It doesn’t matter which point we start at or which direction we proceed in tracing the loop; the voltage sum will still equal zero. To demonstrate, we can tally up the voltages in loop 3-2-1-4-3 of the same circuit:

6.2. KIRCHHOFF’S VOLTAGE LAW (KVL) 177

current

R 1 R 2 R 3

5 kΩ 10 kΩ 7.5 kΩ 45 V

A COM^ V^ Ω

A^ V COM^ Ω A^ V COM^ Ω A^ V COM^ Ω

-20 -15 +

A COM^ V^ Ω

A^ V COM^ Ω

A COM^ V^ Ω

0

-30 V

-45 V

0 V

E3-2 E4-3 E1-4 E2-

E4-

E1-

E2-

The fact that series voltages add up should be no mystery, but we notice that the polarity of these voltages makes a lot of difference in how the figures add. While reading voltage across R 1 , R 1 −−R 2 , and R 1 −−R 2 −−R 3 (I’m using a ”double-dash” symbol ”−−” to represent the series connection between resistors R 1 , R 2 , and R 3 ), we see how the voltages measure successively larger (albeit negative) magnitudes, because the polarities of the individual voltage drops are in the same orientation (positive left, negative right). The sum of the voltage drops across R 1 , R 2 , and R 3 equals 45 volts, which is the same as the battery’s output, except that the battery’s polarity is opposite that of the resistor voltage drops (negative left, positive right), so we end up with 0 volts measured across the whole string of components. That we should end up with exactly 0 volts across the whole string should be no mystery, either. Looking at the circuit, we can see that the far left of the string (left side of R 1 : point number 2) is directly connected to the far right of the string (right side of battery: point number 2), as necessary to complete the circuit. Since these two points are directly connected, they are electrically common to each other. And, as such, the voltage between those two electrically common points must be zero. Kirchhoff’s Voltage Law (sometimes denoted as KVL for short) will work for any circuit config- uration at all, not just simple series. Note how it works for this parallel circuit:

178 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS

6 V^ R 1 R 2 R 3

Being a parallel circuit, the voltage across every resistor is the same as the supply voltage: 6 volts. Tallying up voltages around loop 2-3-4-5-6-7-2, we get:

0 V

voltage from point to point voltage from point to point voltage from point to point

0 V

0 V

voltage from point to point voltage from point to point voltage from point to point

-6 V

0 V

0 V

+6 V

E3-2 =

E4-3 =

E5-4 =

E6-5 =

E7-6 =

E2-7 =

E2-2 =

Note how I label the final (sum) voltage as E 2 − 2. Since we began our loop-stepping sequence at point 2 and ended at point 2, the algebraic sum of those voltages will be the same as the voltage measured between the same point (E 2 − 2 ), which of course must be zero.

The fact that this circuit is parallel instead of series has nothing to do with the validity of Kirchhoff’s Voltage Law. For that matter, the circuit could be a ”black box” – its component configuration completely hidden from our view, with only a set of exposed terminals for us to measure voltage between – and KVL would still hold true:

180 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS

0 V

voltage from point to point voltage from point to point

0 V 3 2

voltage from point to point voltage from point to point

-6 V

0 V

+6 V

E3-2 =

E6-3 =

E3-6 =

E2-3 =

E2-2 =

KVL can be used to determine an unknown voltage in a complex circuit, where all other voltages around a particular ”loop” are known. Take the following complex circuit (actually two series circuits joined by a single wire at the bottom) as an example:

35 V

15 V

20 V

13 V

12 V

25 V

To make the problem simpler, I’ve omitted resistance values and simply given voltage drops across each resistor. The two series circuits share a common wire between them (wire 7-8-9-10), making voltage measurements between the two circuits possible. If we wanted to determine the voltage between points 4 and 3, we could set up a KVL equation with the voltage between those points as the unknown:

E4-3 + E9-4 + E8-9 + E3-8 = 0

E4-3 + 12 + 0 + 20 = 0

E4-3 + 32 = 0

E4-3 = -32 V

6.2. KIRCHHOFF’S VOLTAGE LAW (KVL) 181

1 2

35 V

15 V

20 V

13 V

12 V

25 V

Measuring voltage from point 4 to point 3 (unknown amount)

A^ V COM^ Ω

E4-

???

35 V

15 V

20 V

13 V

12 V

25 V

A^ V COM^ Ω

Measuring voltage from point 9 to point 4 (+12 volts) E4-3 + 12

6.2. KIRCHHOFF’S VOLTAGE LAW (KVL) 183

with the red lead on point 4 and the black lead on point 3:

1 2

35 V

15 V

20 V

13 V

12 V

AV COM^ Ω 25 V

E4-3 = -

In other words, the initial placement of our ”meter leads” in this KVL problem was ”backwards.” Had we generated our KVL equation starting with E 3 − 4 instead of E 4 − 3 , stepping around the same loop with the opposite meter lead orientation, the final answer would have been E 3 − 4 = +32 volts:

1 2

35 V

15 V

20 V

13 V

12 V

AV COM^ Ω 25 V

E3-4 = +

It is important to realize that neither approach is ”wrong.” In both cases, we arrive at the correct assessment of voltage between the two points, 3 and 4: point 3 is positive with respect to point 4, and the voltage between them is 32 volts.

  • REVIEW:
  • Kirchhoff’s Voltage Law (KVL): ”The algebraic sum of all voltages in a loop must equal zero”

184 CHAPTER 6. DIVIDER CIRCUITS AND KIRCHHOFF’S LAWS

6.3 Current divider circuits

Let’s analyze a simple parallel circuit, determining the branch currents through individual resistors:

R 1 R 2 R 3

1 kΩ 3 kΩ 2 kΩ

6 V

Knowing that voltages across all components in a parallel circuit are the same, we can fill in our voltage/current/resistance table with 6 volts across the top row:

E

I

R

Volts Amps Ohms

R 1 R 2 R 3 Total 6 6 6 6

1k 3k 2k Using Ohm’s Law (I=E/R) we can calculate each branch current:

E

I

R

Volts Amps Ohms

R 1 R 2 R 3 Total 6 6 6 6

1k 3k 2k

6m 2m 3m

Knowing that branch currents add up in parallel circuits to equal the total current, we can arrive at total current by summing 6 mA, 2 mA, and 3 mA:

E

I

R

Volts Amps Ohms

R 1 R 2 R 3 Total 6 6 6 6

1k 3k 2k

6m 2m 3m 11m

The final step, of course, is to figure total resistance. This can be done with Ohm’s Law (R=E/I) in the ”total” column, or with the parallel resistance formula from individual resistances. Either way, we’ll get the same answer: