Voltage and Current Division Topic 8 Voltage and Current Division (3.3 & 3.4)   Symbol & MT Extra included

The Voltage Divider v1 + R1 - vs v1 v2 + - Here we apply a source to a couple of resistors in series i By KVL That is, some of the source voltage appears across R1 and some of it across R2. Thus the voltages are given by But how much does each resistor get? The same current i flows through all the resistors & So 4/7/2019 Division

The Voltage Divider R1 R2 vs v1 v2 + - i The voltage is divided proportionally between the two resistors according to their proportion of the total resistance If R1 is v1 is R2 is v2 is of total resistance of vs R1=R2 ½ ½ ½ ½ R2=2R1 ⅓ ⅓ ⅔ ⅔ R1=1KΩ R2=9KΩ 10% 10% 90% 90% 4/7/2019 Division

Loading a Divider A first year Engineering student is asked to get 3 volts out of a 9 volt battery in order to drive a 500Ω load What reading does the student get when the load is connected? Reasoning that a voltage divider should do the trick, the student gets a 1 KΩ resistor and a 2 KΩ resistor then hooks them up as follows: 9v 1KΩ 2KΩ 500Ω 9v 1KΩ 2KΩ v + Hooking a voltmeter across the 1K resistor, the student confirms that there are 3 volts. 4/7/2019 Division

Current Divider Resistors in series give you voltage division + vr + - Resistors in parallel give you current division i1 i2 Now it is the source current which divides to flow through the two resistors So As the resistors are in parallel, they have the same voltage across them, vr Therefore By Ohm’s Law By KCL 4/7/2019 Division

Discussion is R1 R2 i1 i2 vr + - Notice that once again we have a proportionality relationship The more resistance (to flow) the less current that flows through it If R1 is i1 is R2 is i2 is of total resistance of is R1=R2 ½ ½ ½ ½ R2=2R1 ⅓ ⅔ ⅔ ⅓ R1=1KΩ R2=9KΩ 10% 90% 90% 10% 4/7/2019 Division

Water Flow Analogy Big resistance (to flow) = narrower pipe = less flow small resistance (to flow) = wider pipe = more flow 4/7/2019 Division

Current Divider Using Conductance is R1 R2 i1 i2 vr + - Recall that “is defined as” G is conductance which measures the ease of flow Multiply top & bottom by G1G2 Which gives us the same kind of proportionality as exists for voltage dividers and resistance Similarly 4/7/2019 Division

Assesment 3.2 25KΩ (a) Find the no-load value of vo. (no-load means RL is out of there) 75KΩ vo + - RL 200v Using the voltage divider relationship We can just divide out the KΩ So 4/7/2019 Division

Mentally, we have done this Assesment 3.2 25KΩ (b) Find vo when RL is 150KΩ RL is in parallel with the 75KΩ resistor 75KΩ vo + - RL 200v So we can use the voltage divider relationship to write 25KΩ 75KΩ|| RL 200v vo + - Mentally, we have done this So 4/7/2019 Division

General Case of Voltage Division Circuit R1 R2 Rj Rn-1 Rn v + - We have a circuit of some kind that produces a voltage v across its terminals vj + - i We then apply a string of resistors across it in series How much voltage appears across Rj? A single current i flows through the n resistors Thus By KVL So where 4/7/2019 Division

General Case of Voltage Division Circuit R1 R2 Rj Rn-1 Rn v + - vj + - + i vj - j’th resistance voltage across the j’th resistor the original voltage equals over times the equivalent series resistance 4/7/2019 Division

General Case of Current Division Circuit v + - R1 R2 Rj Rn-1 Rn The resistors all experience the same voltage v Some circuit that produces a current i By KCL Has n parallel resistors hooked across it Or where 4/7/2019 Division

General Case of Current Division Circuit v + - ij R1 R2 Rj Rn-1 Rn Then the current through the j’th resistor is Given that Clearly, the voltage across all the resistors is Notice the symmetry with the voltage division case current division voltage division resistors in parallel resistors in series inversely proportional directly proportional 4/7/2019 Division

Which comes first? R1 Both series and parallel Parallel operator is multiplicative, series is additive, so parallel has algebraic precedence Is it R1 in series with R2 first and then R3 in parallel? R2 vo + - R3 vs No, because R1 and R2 do not share the same current. For R1 to be in series with R2 there can be no current split here Does this mean Another way to look at it or It means this 4/7/2019 Division

Assesment 3.4 40Ω 50Ω 70Ω 20Ω 30Ω 10Ω 60V vo + - Use voltage division to determine the voltage vo across the 40Ω resistor a) Now it’s easy 40Ω 70Ω 20Ω 30Ω 60Ω 60V vo + - 40Ω 70Ω 10Ω 60V vo + - 4/7/2019 Division

Assesment 3.4 40Ω 50Ω 70Ω 20Ω 30Ω 10Ω 60V vo + - Use vo to find the current in the 40Ω resistor then use current division to determine the current in the 30Ω resistor b) By Ohm’s Law 40Ω 70Ω 20Ω 30Ω 60Ω 60V 500mA Using our first mental reduction model we can see the current splits three ways. The current through the 30Ω resistor is given by 4/7/2019 Division

The current through the 50Ω resistor Assesment 3.4 40Ω 50Ω 70Ω 20Ω 30Ω 10Ω 60V vo + - i50 How much power is absorbed by the 50Ω resistor c) The current through the 50Ω resistor 40Ω 70Ω 20Ω 30Ω 60Ω 60V 500mA i60 Is the same as the current through the 60Ω resistor in our first reduction Proceeding as in part b) So 4/7/2019 Division

A Perspective on the Parallel Relationship Serial relationship is dominated by the largest resistor For series resistors It should be pretty obvious that Similarly for parallel resistors You could hold R2 steady and vary R1 with the same result Let’s just look at a pair Notice, the combination is always less than either resistor Similarly for R1 Parallel relationship is dominated by the smallest resistor 4/7/2019 Division