Recent content by rms5643

Homework Statement Use the step-by-step method to find vo(t) for t > 0 in the circuit in the figure below. Homework Equations V=IR, KVL, Mesh Analysis, Voltage Division, Solution form of first order equations The Attempt at a Solution Finding the current through the inductor before the...

So, my understanding of voltage division is the following: Suppose you have a circuit with a voltage V and two resistors R1 and R2 shown as follows: --R1-- |...| V...R2 |...| ------- Then, the voltage drop across R1 is defined as follows: Vr1=R1/(R1+R2)*V And the voltage drop across R2 is...

Thank you very much Vela. You were correct. 3mA was incorrect, but 3.05mA was correct. (My online homework has a +/- 2% tolerance, so 0.5 was enough to fall outside of the threshold! Thanks for verifying my methodology.

Homework Statement Find Io in the network in the figure below using linearity and the assumption that Io = 1 mA. Figure:http://i.imgur.com/Xtu0VmG.jpgHomework Equations KCL, KVL, basic analysis techniques.The Attempt at a Solution The following values I have calculated correctly: VR1=9V...

Edit: I solved it, and found that it should be 30v because that causes a 24v drop across 40kohm resistor which is 6v, which is equal to the voltage at the non inverting terminal, meaning Vout will be +VSat, but I don't understand why this is correct, because my notes say that if the voltage at...

Using Ohm's law V=IR, the voltage drop across the 6 ohm resistor is the current, Y, times the resistance, 6 ohms. I am attempting to use W as Vx in the figure and account for the 1/8 in my X mesh equation since it does not appear anywhere else. Would it be more appropriate to assign W=(Y*6/8)?

Thanks for all your help Oxygen, I just noticed I switched up my loop variables when I redid some of the equations, namely ii), which should be 3=Z-Y For iii), that should be W=Y*6 For iv) 6*X+15*X+32*(X-W/8)=0 Now, why is it 32*(X-W/8) and not plus? I defined my W current to go the opposite...

Hey Nascent, that was my error. I don't know why, but I kept ignoring the 15. My equations should be Supermesh: 6*Y-15*X+8*Z=0 Source: 3=X-Y Voltage-controlled current source: W=6*7 Bottom right mesh: 6*X+X+32*(X+W/8)=0 Do these look correct?

Hey again Nascent! Thanks for your help again, my equation -3+V/5-2X+V/4+(V-50)/4=0 you listed as incorrect because of the last term +(V-50)/4. I understand how the current through the 5ohm and 4ohm resistors are flowing downwards (V-0)/5 & (V-0)/4, but I don't understand why (V-50)/4 should be...

Homework Statement Use mesh analysis to find the power delivered by the current-controlled voltage source in the circuit in the Figure: http://i.imgur.com/LUXYFO5.jpg Homework Equations 1 super mesh equation, 1 source equation, 1 mesh equation, KVL The Attempt at a Solution To start off, I...

Thanks for your help everyone! I consider myself to be pretty good with nodal analysis (my preferred method), but I'm really trying to understand mesh analysis because both methods are incredibly powerful. (Note to Electrician: Yes, this is homework for an introductory 200 level E E course. This...

Hey Oxygen, here are my results: KCL @ top node: -3+V/5-2X+V/4+(V-50)/4=0 Relation between V and X: X=(V-50)/4 Solving this gives me that V=-47.5v, multiplying that by the 3A current source gives me 142.5W which is incorrect. Is my KCL equation incorrect or my V/X relationship?

Homework Statement Find the power supplied by the 3-A current source in the network in the Figure using loop analysis: Figure: http://i.imgur.com/OcDUuhX.jpg Homework Equations KVL applied to 1 mesh and 1 super mesh, 1 source equationThe Attempt at a Solution -To start off, I defined all my...

Homework Statement Determine IL in the circuit in the Figure. Figure: http://i.imgur.com/PvKXtjX.png V1 = ? Ix = ? IL = ? 3*Ix current source pointing downwards 7mA current source pointing upwards 7mA current source pointing downwards 7kOhm resistor 1kOhm resistor Homework...

Thanks for your help guys. Since the current would be entering the the negative terminal of R3 into the positive terminal (With respect to the picture), as Nascent said, the voltmeter would read the value as negative. So, the correct answer was -16.058V!