Hello,
if some body know about this question plz help me.THANKS
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Hello,
if some body know about this question plz help me.THANKS
Again your file is corrupted. Can you find a scanner and scan the page in?Quote:
Originally Posted by m777
-Dan
HELLO,
i can attach it again may be it open but i can not scan it.plz try to open it again.Thanks.
Here is the file as a jpg.
RonL
I can figure out what you need using Kirchoff's loop rules, etc, or work it out using equivalent resitances if you think it might help, but I don't know what "nodal analysis" means.
-Dan
I'm confused about something here: Is V0 an applied potential difference or a measured potential difference? If it's an applied potential difference then the equivalent resistance method won't work here. (At least, it won't work in its usual application. I think I can force it to work.)
-Dan
All right. I'll assume that V0 is an input voltage and do it ala Kirchoff's rules.
I'm going to label the intersections of wires to be ABC across the top and DEF across the bottom. (Note: I am ignoring the ammeter part of the circuit.)
So, we need to assign directions for the currents. I don't know if you are in EE or Physics. I'll assume EE and put the current as going from the source to point A and call it Ieq. (I'm labelling currents and potentials in reference to which resistance they cross. Ieq crosses the equivalent resistance of the circuit, hence I eq.) I've got I6 going from A to D, I3 going from A to B, I10 going from B to E, I4 going from B to C, and I2 going from C to F.
The junction rule says that the current going into an intersection is equal to the net current leaving it. Thus:
Ieq = I6 + I3
I3 = I10 + I4
I4 = I2 (The resistors are in series)
We have 4 unknowns and two equations (using the I4 = I2 identity. We also know the value of I3: I3 = 4 mA according to the diagram.) That means we need 2 more equations.
Loop ABDE:
The loop rule says that the potential drop across a connected loop of circuit is 0 V. So starting at point A we cross the 3 kOhm resistor and since the current is in the direction we are going around the loop (clockwise in this case) we drop the potential by I3*R3 = (4 mA)*(3 kOhm) = 12 V. Next we meet the 10 kOhm resistor and drop another I10*R10 = 10000*(I10). (Remember the unit is kOhm!) Finally we cross the 6 kOhm resistor in the direction opposite to the current flow we chose. So we raise the potential by I6*R6 = 6000*(I6). Then we're back to point A. So our loop equation is:
-12 - 10000*(I10) + 6000*(I6) = 0
Loop BCEF:
This loop is similar in all details to the previous loop, except with the reminder that I2 = I4. So the loop equation is:
-4000*(I4) - 2000*(I4) + 10000*(I10) = 0
Noting that V0 = I4*R2 and VI = I10*R10 gives
V0 = 2000*(I4)
VI = 10000*(I10)
So we need I4 and I10 specifically.
Our equations are now:
Ieq = I6 + 4 x 10^{-3}
4 x 10^{-3}= I10 + I4
-12 - 10000*(I10) + 6000*(I6) = 0
-4000*(I4) - 2000*(I4) + 10000*(I10) = 0
I'm going to make a simplification here and I hope it doesn't confuse you. I'm going to assume the units for the current are mA, meaning that our equations will generate "nicer looking" numbers:
Ieq = I6 + 4
4 = I10 + I4
-12 - 10*(I10) + 6*(I6) = 0
-4*(I4) - 2*(I4) + 10*(I10) = 0
All we need from this are values of I4 and I10. If you look at the equations there are two equations that deal soley in only the variables I4 and I10:
4 = I10 + I4
-4*(I4) - 2*(I4) + 10*(I10) = 0
So all we need do is solve these. Simplifying the second equation a bit:
4 = I10 + I4
-6*(I4) + 10*(I10) = 0
The first equation says:
I10 = 4 - I4
Inserting this into the second equation:
-6*(I4) + 10*(4 - I4) = 0
-6*(I4) + 40 - 10*(I4) = 0
-16*(I4) = -40
I4 = -40/-16 = 5/2 mA = 2.5 mA (Recall that we assumed the units for currents were in mA.)
Then
I10 = 4 - I4 = 4 - 5/2 = 3/2 mA = 1.5 mA.
So:
V0 = 2000*(I4) = 2000*(2.5 x 10^{-3}) = 5 V
VI = 10000*(I10) = 10000*(1.5 x 10^{-3}) = 15 V.
-Dan