How to use Thevenin to find R and I?

  • #1

Homework Statement


In the circuit shown, in the image, the variable resistor R is consuming maximum power. Determine:
a) resistance for R
b) power for R
c) current for R3 and its direction
uMHqpqA.png

Homework Equations




The Attempt at a Solution


I tried getting IA(the direction is shown with arrows in the picture under) like this: E1-IAR3+IR3-IAR4=0
-5IA-5IA=30-10
IA=-2 A
but I have a feeling that's not good ^
---------------------------------------------------------------------------------------------------------------------------------
wsuU2b9.png

I calculated RT using Thevenin: RT=Rab=R3+R4=10Ω
Then I got from the main scheme:
Uab=-IAR4+E1-IAR3+IR3 = -(-2)*5+30-(-2)*5+10 = 60 V

I don't know how to proceed now.
 

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Answers and Replies

  • #2
gneill
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Investigate the "Maximum power transfer theorem".
 
  • #3
Investigate the "Maximum power transfer theorem".
So according to the "maximum power transfer theorem", R=RT=10Ω. Thank you.

From that, the equation for IA should be:
E1-IAR3+IR3-IAR-IR-IAR4=0
-20IA=-30-10-20
IA=3 A
and for Uab:
Uab=-IAR4+E1-IAR3+IR3=40 V
Now I can get IR=ET/RT+R=2 A and PR=I2*R=22*10=40 W. Hopefully I did everything right.

Although, now it asks me to get IR3 and its direction, but isn't its direction already defined by the current source?
 
  • #4
Maybe it's: IR3=I+I-IA ? With direction to the left?
 
  • #5
gneill
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From that, the equation for IA should be:
E1-IAR3+IR3-IAR-IR-IAR4=0
-20IA=-30-10-20
IA=3 A
and for Uab:
Uab=-IAR4+E1-IAR3+IR3=40 V
Now I can get IR=ET/RT+R=2 A and PR=I2*R=22*10=40 W. Hopefully I did everything right.
Looks good. A simpler way would be to employ your Thevenin equivalent:
upload_2018-12-13_12-43-28.png


##I = \frac{V_{th}}{R_{th} + R_L}##

##V_{ab} = V_{th} \frac{R_L}{R_{th} + R_L}##
Although, now it asks me to get IR3 and its direction, but isn't its direction already defined by the current source?
Not quite, since neither of the current sources is directly in series with ##R_3##, and you may not be able to know off hand the effect of ##E_1## on the direction. However, you do now know all the currents but one entering/leaving the central node, so KCL at that node will settle the matter.
Maybe it's: IR3=I+I-IA ? With direction to the left?
Yup.
 

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Last edited:
  • #6
Looks good. A simpler way would be to employ your Thevenin equivalent: Vab=Vth*RL/Rth+RL
This gives me Vab = 20V, but my way above got Vab = 40V?
 
  • #7
gneill
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This gives me Vab = 20V, but my way above got Vab = 40V?
Oops, I missed that when I scanned your work. Although you shouldn't be getting 20 V either !

By the Thevenin method you know you've got a voltage divider with two equal resistances, hence the output should be 1/2 the source voltage, so one half ##V_{th}## or 30 V.
 
  • #8
Oops, I missed that when I scanned your work. Although you shouldn't be getting 20 V either !

By the Thevenin method you know you've got a voltage divider with two equal resistances, hence the output should be 1/2 the source voltage, so one half ##V_{th}## or 30 V.
I am confused now. Isn't the Vab I got also Vth? Where did you get 30 V from?
 
  • #9
gneill
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I am confused now. Isn't the Vab I got also Vth? Where did you get 30 V from?
See the diagram in post #5.

You calculated ##V_{th} = 60\;V## and ##R_{th} = 10\;\Omega##. Half of ##V_{th}## is ##30\;V##.
 
  • #10
See the diagram in post #5.

You calculated ##V_{th} = 60\;V## and ##R_{th} = 10\;\Omega##. Half of ##V_{th}## is ##30\;V##.
It says 40V for me, not 60V?
VbFI78b.png

I also thought Vab = Vth?
 

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  • #11
gneill
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In your very first post you correctly calculated the open circuit voltage ##U_{ab}## which is the Thevenin voltage. Once you attach a load across the output, it will no longer read ##V_{th}## but some fraction of it depending on the size of the load resistance and ##R_{th}##.
 
  • #12
In your very first post you correctly calculated the open circuit voltage ##U_{ab}## which is the Thevenin voltage. Once you attach a load across the output, it will no longer read ##V_{th}## but some fraction of it depending on the size of the load resistance and ##R_{th}##.
Alright, but how does that fit into the maximum power transfer theorem if RT=R=10Ω then gives me IA=3 A and with that IA I get Vab=40V?
Because the first time I solved for IA, I did it like R wasn't there at all, but since RT=R=10Ω, shouldn't I solve it using R in the equation as well?
 
  • #13
gneill
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Alright, but how does that fit into the maximum power transfer theorem if RT=R=10Ω then gives me IA=3 A and with that IA I get Vab=40V?
Because the first time I solved for IA, I did it like R wasn't there at all, but since RT=R=10Ω, shouldn't I solve it using R in the equation as well?
Yes. For the Thevenin version you've two resistors forming a voltage divider. Each of them is 10 Ω. Also,

3 A * 10 Ω = 30 V.
 
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  • #14
Okay I think I kind of understand it now. Thank you.
 

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