simple circuit problems

projection

Joined Oct 17, 2009
6
I am having trouble solving these two problems... some help would be greatly appreciated.

#1 - Find the voltage of v1.

I am unsure what method to use, nodal analysis or (KVL and KCL) combination. i don't know that to do with the 4V at the end... how do i incorporate that into the problem .

( on a side note, what does it mean when the voltage (4V) is given like that... is that a component/circuit element or is it the voltage of the branch? Are the 1 Ohm and 2 Ohm on the right in series or the little branch that is going to the 4 V + side affecting it? Really confused about the presence of the 4V and the meaning of the small extensions.)

#2 - Find the power power delivered by the dependent voltage source.

For this one I think nodal analysis it the way to go but i don't know if i should add the two resistances (1.8 and 4.7) connected to the 200(Idelta) source when i am writing the KCL for the node... some clarification on what steps to take would be nice.

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hgmjr

Joined Jan 28, 2005
9,029
It appears that problem #1 is giving the output voltage so that the goal must be to determine what voltage V1 would yield 4 Volts.

hgmjr

projection

Joined Oct 17, 2009
6
It appears that problem #1 is giving the output voltage so that the goal must be to determine what voltage V1 would yield 4 Volts.

hgmjr
can you explain what the role of the 4V is... output, as in that it is just there to inform about the voltage of the circuit or is it part of the circuit... i looked every where for information about the description of the way it is drawn.

hgmjr

Joined Jan 28, 2005
9,029
All the 4 Volts is indicating is that with a voltmeter connected to the + and - terminals, you would measure 4 Volts.

hgmjr

projection

Joined Oct 17, 2009
6
All the 4 Volts is indicating is that with a voltmeter connected to the + and - terminals, you would measure 4 Volts.

hgmjr
thanks, that helps... i was under the impression it was part of the circuit... this would mean the the current through the 2 ohm and 1 ohm resistor is the same... that is 3 A... then since the voltage of the the entire branch is 4V... getting the voltage of the 1 ohm resistor by ohm's law (getting 3V). We can find the voltage of the middle 2 ohm resistor by KVL... then applying KCL to the node would get us the current through the last 2 ohm resistor... now should the final KVL be applied to the first through branches or the large one around the circuit?

can you provided some assistance with #2... do i combine the 4.7 and 1.8 resistors?

hgmjr

Joined Jan 28, 2005
9,029
thanks, that helps... i was under the impression it was part of the circuit... this would mean the the current through the 2 ohm and 1 ohm resistor is the same... that is 3 A... then since the voltage of the the entire branch is 4V... getting the voltage of the 1 ohm resistor by ohm's law (getting 3V). We can find the voltage of the middle 2 ohm resistor by KVL... then applying KCL to the node would get us the current through the last 2 ohm resistor... now should the final KVL be applied to the first through branches or the large one around the circuit?
Looks like you are well on your way to solving #1.

I haven't looked too deeply and #2 yet.

hgmjr

hgmjr

Joined Jan 28, 2005
9,029
can you provided some assistance with #2... do i combine the 4.7 and 1.8 resistors?
YES....

hgmjr

projection

Joined Oct 17, 2009
6
ok. for #1 i get the right answer when i apply kvl to the first loop (between the 2 ohm branches) but get a a very different answer when i apply it to the loop around the whole circuit... is this possible?

hgmjr

Joined Jan 28, 2005
9,029
ok. for #1 i get the right answer when i apply kvl to the first loop (between the 2 ohm branches) but get a a very different answer when i apply it to the loop around the whole circuit... is this possible?
I suspect you have a hiccup in your problem setup if you get the wrong answer. I will need to see your work to get to the bottom of this paradox.

hgmjr

projection

Joined Oct 17, 2009
6
for #2, i tried node analysis at the top:

-0.008 - v1/3300 + (v1 - 200 Idelta)/6500 = 0

Idelta = -v1/3300 , therefore v1 = -57.2 V

I delta = 0.017333

Voltage at dependent source is V = 0.017333(200) = 3.467 V

I at dependent voltage source is : (v1 - 200Idelta) / 6500 = -0.009333 A

p=VI = (3.467)(-0.009333) = -32.36 mW ... so thats 32.36 mW delievered... but the answer is 2.87mW. what am i doing wrong?

hgmjr

Joined Jan 28, 2005
9,029
$-0.008-\frac{V1}{3300}+\frac{V1-200i}{6500}=0$

$i=-\frac{V1}{3300}$
My equations:

$0.008-\frac{V1}{3300}- \left[\frac{V1-200i}{6500}\right]=0$

$i=\frac{V1}{3300}$
I get 2.79 millWatts.

hgmjr

projection

Joined Oct 17, 2009
6
thanks.

so the reason you have it as -v1/3300 (when you made it positive entering the node) is because Idelta is going from negative to positive in reference to the node voltage???

i ended up getting -2.9mW... close to the 2.87mW answer. (i'm assuming they dropped the negative because it is negative by stating DELIVERED.)

hgmjr

Joined Jan 28, 2005
9,029
What was the voltage you got at the junction of the 980 ohm, 3.3K, and 1.8K resistors?

hgmjr

Ratch

Joined Mar 20, 2007
1,068
Here is what I got. The right loop equation is, assuming current direction is clockwise on both loops, and using KVL counterclockwise.

-3.3K(0.008-I)+I*6.5K-(0.008-I)*200=0 ===> I = 2.8ma

So 2.8ma current exists in a clockwise direction in the right loop. That works out to (0.008-I)*3300 = 17.16volts across the 3.3k resistor. Now the power supplied by the dependent voltage source is 0.0028*(0.008-.0028)*200 = 2.91mw. The power dissipated in the circuit by the dependent voltage source is 6500*0.0028^2-17.16*0.0028 = 2.91mw. The minus sign in the above equation is because the 8ma current source is supplying power to the dependent voltage source. I will be glad to explain my reasoning for that statement if necessary.

Ratch

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