Find current using KVL (with dependent source)

AI Thread Summary
The discussion focuses on applying Kirchhoff's Voltage Law (KVL) to find the current using dependent sources in a circuit. The user initially calculated the current Ix as 2 mA but expected it to be 1.5 mA, prompting a request for help in identifying the error. After reviewing the equations, it was determined that the mistake was in not properly accounting for the unit conversion of Ix from mA. The corrected equations were confirmed, leading to the resolution of the problem. This highlights the importance of unit consistency in circuit analysis.
iharuyuki
Messages
15
Reaction score
1

Homework Statement


upload_2015-4-15_22-24-13.png


Homework Equations


V = IR
Kirchoff's Voltage Law

The Attempt at a Solution


[/B]
Top loop - a
Bottom right - b
Bottom left - c
Ix = x

KVL:
top: a(6000) + (a-b)(2000) + 5x - 10=0
bottom right: -5x + (b-a)(2000) + b(3000) + (b-c)(4000) = 0
bottom left : 10 + (c-b)(4000) + c(2000) = 0
x = -c

x comes out to 2 mA when it should be 1.5. Where did I make a mistake?

Thank you very much.
 
Physics news on Phys.org
How did you account for the given detail that ##I_x## is in mA?
 
got it! That was the error. The proper equations were this:

a(6000) + (a-b)(2000) + 5x - 10=0

-5x + (b-a)(2000) + b(3000) + (b-c)(4000) = 0

10 + (c-b)(4000) + c(2000) = 0

x/1000 = -c

Thank you very much!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top