Kirchoffs law question - Different values!?
Answer:
I suggest you apply the nodal voltage method for circuit analysis:
(V-80)/(80+130) + (V-105)/(105) + (V-0)/(175+130) = 0
Solve for V.
Note how the resistors in each branch are added.
Note how the current in each branch is given implicitly as a term.
For the explicit current in each branch, solve for V and then evaluate the appropriate terms explicitly.
The problem reduces to a fairly simple (but not a trivial integer) solution.
Good luck!
I see a scanned-in homework problem. What's your question? I'm not going to simply solve the circuit for you -- that would be cheating.
Let us know where you're stuck, and we'll give you a hand. Yes, Kirchoff's law applies to this circuit.
Review the section in your text on nodal equations and mesh analysis. Much more useful and satisfying in the long run than having us cheat for you!
use the method of superposition in which u will consider E3 existing and E4 short circuited then make E4 existing and E3 short circuited
make superimposed solution
u will get fast and easy
Assume direction of currents:
I1= flows ccw in loop 1, E3
I2=flows ccw in loop 2, E4
I3=I1-I2, flowing thru E4, minus because they oppose each other
By Kirchoffs:
Equation at loop 1;
-80+80(I1)+105(I1)-105(I2)+105...
315(I1)=-25+105(I2)
I1=-25/315 +105(I2)/315=-0.079+0.33I2
Equation at loop 2;
-105+105(I2)+130(I2)+175(I2)-1...
410(I2)=105+105(I1)
I2=105/410 +105(I1)/410=0.256+0.256(I1)
Subst. I2 in first equation to find I1;
I1=-0.079+0.33*(0.256+0.256I1)
I1=-0.079+0.084+0.084I1
I1-0.084(I1)=0.005
0.916(I1)=0.005
I1=0.00545amps flowing thru E3
Subst. I1 in equation 2 to get I2:
I2=0.256+0.256(I1)
=0.256+0.256*0.00545=0.256+0.0...
I3=I2-I1=0.257-0.00545=0.2519 amps, flowing thru E4
Iload=I2 @130 ohm = 0.257 amps
Pload= I^2R=(0.257)^2*130=8.56 watts
answers:
a)I flows @E3=I1=0.00545 amps
b) I flows @E4=I3= 0.2519 amps
c) I load @130 ohms=I2= 0.257 amps
d) P dissipated @ 130 ohms=8.56 watts
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