Transcript Voltage and Current Division
Circuits with Dependent Sources Chapter 3
Circuit with Dependent Sources V 1 = 60 volts because the 20 parallel; by Ohm’s law, V 1 Ω resistor is in = I 2 ·20Ω; so I 2 = V 1 /20 Ω = 60v/20Ω = 3 A
Circuit with Dependent Sources If I 2 = 3 A, then the 5 Ω·I 2 dependent source is 15 volts and if V 1 = 60 v., then the V 1 /4 Ω dependent source is 15 A
Circuit with Dependent Sources Writing Kirchoff’s Voltage law around the outside loop, -60 v + 5 Ω·I 2 where I 2 =3 A, so I 3 + 5 Ω·I 3 = 0 = (60 –15)v / 5Ω = 9 A
Circuit with Dependent Sources Writing Kirchoff’s Current law at B I 4 + I 3 + V 1 /4 = 0 (all leaving node B) Since V 1 /4 Ω =15 A and I 3 = 9 A, I 4 = -24 A
Circuit with Dependent Sources Writing Kirchoff’s Current law at A I 4 + I 1 – I 2 = 0 Since I 2 = 3 A and I 4 = -24 A, I 1 = 27 A
Circuit with Dependent Sources
I 4 = -24A I 1 =27A I 1 /4Ω=15A V 2 = 45v
60 v source generating, P=-27A ·60v=-1620 watts 5 ·I 2 source absorbing, P=24A ·15v=360 watts V 1 /4 source absorbing, P=15A ·45v= 675watts
Circuit with Dependent Sources
I 3 =9A V 1 =60v I 2 =3A
20 Ω resistor absorbing, P=3A·60v=180 watts 5 Ω resistor absorbing, P=9A·45v= 405 watts -1620 w +360 w + 675 w + 180 w + 405 w = 0
V 2 =45v
VA = 28 volts Find I 1 Find I 2 Find V B Find I 3 Find I 4 2 Amps by Ohm’s law (I 2 ∙14=28) 2 Amps by KCL (4-I 1 -I 2 =0) 24 volts by KVL (-I 1 ∙14+I 2 ∙2+V B =0) 6 Amps by Ohm’s law (I 3 ∙4=24) 4 Amps by KCL (I 2 +I 4 -I 3 =0)
Sources in Series + V1 + V2 + V1+V2
Voltage sources In series add algebraically
Sources in Series + V1 + V2 + V1-V2
Start at the top terminal and add.
If hit a + (+V1) If you hit a – (-V2)
Sources in Series + 8 ·Vx + V2 + 8 ·Vx-V2
If one source is dependent, then so is the equivalent
Is Is Sources in Series Is
Current sources in series must be the same value and direction
Sources in Parallel
Current sources in parallel add algebraically
I1+I2 I1 I2
Sources in Parallel
Current sources in parallel add algebraically
-I1+I2 I1 I2
I1 Sources in Parallel
If any source is dependent, then the combination is also dependent
5 ·Ix -I1+5 ·Ix
Sources in Parallel
Voltage sources in parallel must be the same value and same direction
+ Vs + Vs + Vs
Source Transformation Section 5.2
Source Transformation • Practical voltage sources are current limited and we can model them by adding a resistor in series
Practical Source
R S + I L + Vs R L V L • We want to create an equivalent using a current source and parallel resistance for any R L
Source Transformation • V L and I L for any R L must be the same in both circuits
Practical Source
+ I L Ip R p R L V L -
Source Transformation • V L and I L must be the same in both circuits for R L = 0, or a short circuit
Practical Source
+ I L Ip R p R L =0 V L • Ip = I L and V L = 0
Source Transformation • Now look at the voltage source in series with the resistor with a short circuit
Practical Source
R S + I L + Vs R L =0 V L • I L = Vs/Rs and V L • So Ip = Vs/Rs = 0
Source Transformation • V L and I L must also be the same in both circuits for R L = ∞, or an open circuit
Practical Source
+ I L Ip R p R L = ∞ V L • I L = 0 and V L = Ip ·Rp
Source Transformation • Now look at the voltage source in series with the resistor with an open circuit
Practical Source
R S + I L + Vs R L = ∞ V L • I L = 0 and V L = Vs, so Vs = Ip • If Ip = Vs/Rs, then Rp = Rs ·Rp
Example • We can transform the voltage source + 20v 10 Ω
I 1
15 Ω 6 Ω 4A • Why? Gets all components in parallel + Vo
Example • We can combine sources and resistors 20v/10 Ω =2A 10 Ω 15 Ω 6 Ω • Ieq = 2A+4A = 6A, Req = 3Ω 4A + Vo -
Example • Vo = 6A· 3Ω = 18 v 6A + 3 Ω Vo -
Example • Going back to the original circuit, Vo=18 v + 20v 10 Ω
I 1
15 Ω
18/15 =1.2A
6 Ω
18/6 =3A
4A + Vo = 18 v • KCL: I 1 + 4A - 1.2A - 3A=0, so I 1 =0.2A
-
Example • We can transform the current source after first combining parallel resistances 10 Ω + 4A + 20v 15 Ω 6 Ω Vo • Why? Gets all components in series -
Example • We can transform the current source after first combining parallel resistances 10 Ω + 4A + 20v 30/7 Ω Vo • Req=6·15/(6+15)=30/7 Ω
Example • We can now add the series voltage sources and resistances 10 Ω 30/7 Ω
I 1
+ 20v 120/7v + • Rtotal=100/7 Ω and Vtotal=20/7 volts
Example • We can easily solve using KVL for I 1 100/7 Ω
I 1
+ 20/7 v • I 1 = 20/7 ÷ 100/7 = 0.2 A
Voltage & Current Division Chapter 3, Section 8
Voltage Division • We could have a circuit with multiple resistors in series where we want to be able to find the voltage across any resistor
. . .
. . .
R1 Ri
+ Vi I
+ Vs Rn • Clearly Req = Σ Ri, and I = Vs/Req • So Vi = I·Ri = Vs ·(Ri/Req)
Voltage Division Application • You have a 12 volt source, but some devices in your circuit need voltages of 3 and 9 volts to run properly
. . .
. . .
R1 Ri
+ Vi I
+ Vs Rn • You can design a voltage divider circuit to produce the necessary voltages
Voltage Division Application • To get 3, 6 and 3 across the three resistors, any R, 2 ·R and R could be used + -
12 v
R
+ 3v + 9v A
2 ·R
+ 6v -
• 9 volts is available at A, 3 volts at B R
+ B 3v
Wheatstone Bridge • This circuit is often used to measure resistance or convert resistance into a voltage.
300Ω 100Ω + 100v A + V AB B 500Ω R
Wheatstone Bridge • Using the voltage divider at A, • V AD = 100 v ∙ R/(100+R)Ω
+ 100v A 100Ω + V AB 300Ω B R 500Ω D
Wheatstone Bridge • Find the Voltage at B, using the voltage divider theorem
+ 100v A 100Ω + V AB 300Ω B R 500Ω
Wheatstone Bridge • V BD = 100 v ∙ 500Ω/(300+500)Ω = 62.5 v
+ 100v A 100Ω + V AB 300Ω B R 500Ω D
Wheatstone Bridge • Let’s find the relationship between V AB V AB = V AD – V BD = 100∙R/(100+R) – 62.5
& R
+ 100v A 100Ω + V AB 300Ω B R 500Ω D
Wheatstone Bridge • The Wheatstone bridge is considered balanced when V AB =0 v.
• Find R • R=167Ω
100Ω 300Ω + 100v A + V AB B 500Ω R
Wheatstone Bridge
Current Division • What if we want to find the current through any parallel resistor?
. . .
. . .
+ Is R1 Ri
Ii
Rn V • Req = 1 / Σ(1/Ri) and V = Is·Req • So Ii = V / Ri = Is·(Req/Ri) -
Wheatstone Bridge • The 10 A current source divides between the two branches of the bridge circuit
10 A I 1 100Ω + V AB 300Ω I 2 100Ω 500Ω D
Wheatstone Bridge • First, simplify by combining the series resistances in each branch of the bridge
10 A I 1 100Ω + V AB 300Ω I 2 100Ω 500Ω D
Wheatstone Bridge • First, simplify by combining the series resistances in each branch of the bridge
10 A I 1 100+100Ω I 2 300+500Ω
Wheatstone Bridge • Find the parallel equivalent resistance • Req = 200∙800/(200+800) Ω = 160 Ω
10 A I 1 100+100Ω I 2 300+500Ω
Wheatstone Bridge • I 1 • I 2 = 10 A ∙(160/200) = 8 A = 10 A ∙(160/800) = 2 A
I 1 I 2 10 A 100+100Ω 300+500Ω
Voltage vs Current Division • Voltage Division • Current Division • Vi = Vs · Ri/Req • Req = Σ Ri • Ri/Req < 1 • Vi < Vs • Series resistors only • Ii = Is · Req/Ri • Req = 1÷ Σ (1/Ri) • Req/Ri < 1 • Ii < Is • Parallel resistors only
Application • A practical source of 12 volts has a 1Ω internal resistance. Design a voltage divider that will be used to power up to ten 6 volt light bulbs in parallel.
1Ω 12 v + R 1 R 2 100Ω . . . .
10 lights 100Ω 100Ω
Application • The voltage across each bulb must be within the range of 5.5 to 6 volts in order for the bulbs to be bright enough. Find R 1 and R 2 .
1Ω 12 v + R 1 R 2 100Ω . . . .
10 lights 100Ω 100Ω
Application • With no bulbs, the maximum voltage is obtained:
Vbulb = 6.0 v =
12v·R 2 1 Ω+R 1 +R 2
1Ω 12 v + R 1 R 2 + Vbulb
Application • The parallel combination of ten 100Ω bulbs is 10 Ω, the parallel combination with R2 is 10·R 2 /(10+R 2 )
Ω
10·R 2 /(10+R 2 )
1Ω 12 v + R 1 R 2 100Ω . . . .
10 lights 100Ω 100Ω + Vbulb
Application • Using the voltage divider and the minimum voltage allowed:
Vbulb = 5.5 v =
12v·10·R 2 /(10+R 2 ) 1 Ω+R 1 +10·R 2 /(10+R 2 )
1Ω 12 v + R 1 R 2 100Ω . . . .
10 lights 100Ω 100Ω + Vbulb
Solution • Solving the two equations simultaneously, R 1 = .82
Ω and R 2 = 1.82
Ω • With 10 bulbs, R 2 in parallel with ten 100 resistors would be 1.54
Ω Ω • Using the voltage divider, with 10 bulbs Vbulb = 12∙1.54Ω÷(1+.82Ω+1.54Ω) = 5.5 v • Using the voltage divider, with no bulbs Vbulb = 12∙1.82Ω÷(1+.82Ω+1.82Ω) = 6.0 v