Chapter 4 Series Circuits

Chapter 4Series Circuits

ElectronicCircuits&Electronics Shao-YuLien 1

4.1 Resistors in Series

Definition. A series circuit provides only one path for current between two points so that the current is the same through each series device


4.2 Total Series Resistance

• When resistors are connected in series, the resistor values addbecause each resistor offers opposition to the current in direct proportion to current



• Series resistance formula: For any number of individual resistors connected in series, the total resistance is the sum of each of the individual values

RT=R1+R2+R3+…+Rn

– RT: Total resistance– n: Any positive integer



• RT=56 Ω+100 Ω+27 Ω+10 Ω+47 Ω=240 Ω



Example. Calculate the total resistance RT for each circuit

• For (a), RT=1.0 kΩ+2.2 kΩ+4.7 kΩ+10 kΩ=17.9 kΩ• For (b), RT=4.7 kΩ+10 kΩ+1.0 kΩ+2.2 kΩ=17.9 kΩ


4.3 Current in a Series Circuit

• At any point in this circuit, the current into that point must equal the current out of this point


4.4 Application of Ohm’s Law

• Current through any of the series resistors is the same as the total current

• If you know that total applied voltage and the total resistance, you can determine the total current by Ohm’s law

𝐼# =𝑉#𝑅#

• If you know the voltage drop across one of the series resistors,Rx, you can determine the total current by Ohm’s law

𝐼# =𝑉'𝑅'



• If you know the total current, you can find the voltage drop across any of the series resistors by Ohm’s law

Vx=ITRx

• The polarity of a voltage drop across a resistor is positive at the end of the resistor that is closest to the positive terminal of the voltage source

• The current through a resistor is defined to be in a direction from the negative of the resistor to the positive end

• An open in a series circuit prevents current; and therefore, there is zero voltage drop across each series resistor. The total voltage appears across the points between which there is an open



Example. Calculate the voltage drop across each resistor in the following figure, and find the value of VS. To what value can VS be raised if the current is to be limited to 5 mA?



• The voltage drop across R1 is V1=IR1=(1 mA)(1.0 kΩ)=1 V

• The voltage drop across R2 isV2=IR2=(1 mA)(3.3 kΩ)=3.3 V

• The voltage drop across R3 isV2=IR2=(1 mA)(4.7 kΩ)=4.7 V

• To calculate the value of VS, first determine the total resistanceRT=1.0 kΩ+3.3 kΩ+4.7 kΩ=9 kΩ

• The source voltage VS=IRT=(1 mA)(9 kΩ)=9 V• When I=5 mA, VS can be increased to the maximum value

VS(max)=IRT==(5 mA)(9 kΩ)=45 VElectronicCircuits&Electronics Shao-YuLien 11

4.5 Voltage Sources in Series

• Series voltage sources are added when their polarities are in the same direction, and are subtracted when their polarities are in opposite direction


4.5 Voltage Sources in Series

Example. What is the total source voltage (VS(tot)) in following figure.

• The polarity of each source is the same. Therefore, the total voltage is

VS(tot)=6 V+6V+6V=18 VElectronicCircuits&Electronics Shao-YuLien 13

4.6 Kirchhoff’s Voltage

• Kirchhoff’s voltage law: The sum of all voltage drop around a single closed path in a circuit is equal to the total source voltage in that closed path.

VS=V1+V2+…+Vn



• Kirchhoff’s voltage law was developed for a series of circuit,but the concept can be applied to any circuit

• You can still write Kirchhoff’s voltage law, but there are caseswhere there is no source voltage in a given closed loop

• The algebraic sum of the voltage around any closed path ina circuit is equal to zero

V1+V2+…+Vn=0– Any of the variables above can represent a voltage rise or a voltage

drop



Example. Determine the unknown voltage drop V3 in the following figure

V1+V2+V3-VS=0V3=VS-V1-V2=13 V


4.7 Voltage Divider

• A circuit consisting of a series string of resistors connected to a voltage source acts as a voltage divider

• The total voltage drop around a single closed path divides among the series resistors in amounts directly proportional to the resistance values


4.7 Voltage Divider


• Let Vx represent the voltage drop across any one of the resistor• Let Rx represent the value of a particular resistor

4.7 Voltage Divider


Vx=IRx

• The total resistance isRT=R1+R2+R3+…+Rn

and the current isI=VS/RT

• By substitution of VS/RT for I in the expression of Vx

𝑉' =𝑉(𝑅#

𝑅'

• Rearranging the terms you get

𝑉' =)*)+

𝑉,

4.7 Voltage Divider

• The voltage drop across any resistor or combination of resistors in a series circuit is equal to the ratio of that resistance voltage value to the total resistance, multiplied by the source voltage


4.7 Voltage Divider

• Determine the voltage between the following points in the circuit

– A to B– A to C– B to C– B to D


4.7 Voltage Divider

• First determine RT

RT=1.0 kΩ+8.2 kΩ+3.3 kΩ=12.5 kΩ• The voltage A to B is the voltage drop across R1

𝑉-. =)/)+

𝑉, =0.24506.745

25 V=2 V

• The voltage from A to C is the combined drop across both R1and R2

𝑉-8 =)/9):)+

𝑉, =;.64506.745

25 V=18.4 V

• The voltage B to C is the voltage drop across R2

𝑉.8 =):)+

𝑉, =<.64506.745

25 V=16.4 V


4.7 Voltage Divider

• The voltage from B to D is the combined drop across both R2and R3

𝑉.= =):9)>)+

𝑉, =00.74506.745

25 V=23 V


4.7 Voltage Divider

• The potentiometer as an adjustable voltage divider


4.8 Power in Series Circuits

• The total amount of power in a series resistive circuit is equal to the sum of the power in each resistor in series

PT=P1+P2+…+Pn– PT: Total power– n: Number of resistors

• The following formulas are used to calculate the total powerPT=VSIPT=I2RT

𝑃# =𝑉,6

𝑅#



Example. Determine the total amount of power in the series circuit and the power of each resistor



• The source voltage is 15 V. The total resistance is RT=10 Ω+12 Ω+56 Ω+22 Ω=100 Ω

• We can use PT=𝑉,6/RT

PT=(15 V)2/100 Ω=2.25 W• Find the current

I=VS/RT=(15 V)/(100 Ω)=150 mA• Calculate the power for each resistor P=I2R

P1=(150 mA)2(10 Ω)=225 mWP2=(150 mA)2(12 Ω)=270 mWP3=(150 mA)2(56 Ω)=1.26 WP4=(150 mA)2(22 Ω)=495 mW



• The total powerPT=225 mA+270 mA+1.26 W+495 mW=2.25 W


4.9 Voltage Measurements

• Measuring voltage with respect to the ground


Chapter 4 Series Circuits

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