In this question use Mash anlysis to find voltage?

Question

pelase see http://sit.edu.pk/question/Mesh%20Analysis_Q2.doc

Answer 1

I will work in mA and volts.
Start with the right loop
2*I1-50*Ix-4*I2=0
(Vo=4*I2)

The left loop
10-6*Ix-2*I1=0

Summing currents at the bottom node
Ix-I1-I2=0

This gives the following equations
I1 I2 Ix Constant
2 -4 -50 0
-2 0 -6 10
-1 -1 1 0
The following coefficients solve for Ix
I1 I2 Ix Constant
2 -4 -50 0
-6 0 -18 30
4 4 -4 0

0 0 -72 30



72*Ix=30
Ix=5/12 mA
Solving for I1
6*I1+18*5/12=30

I1=(30-18*5/12)/6
I1=3.75 mA

I2=5/12-3.75
I2=-2 mA

Vo=-8V

j

Answer 2

This is similar to the other problem, but you have to use a different technique. Once again, it's hard to explain without drawing.

Let your variables be I1, I2, and I3. I1 is the left loop, I2 the middle loop, and I3 the right loop (open to the battery). For consistency, draw all the currents clockwise.

Now apply the loop law. Voltage around the loop is zero.

Applying it to the left loop:

10V - (6 kiloohms)I1 - (2 kiloohms)(I1 - I2) = 0

The I1 - I2 is because loop 1 and 2 both contribute to the current through that resistor.

Do the same thing around your other loops.

This gives you 3 equations and 3 unknowns. Do algebra to find the currents. Once you have all the currents, it's easy to find the voltages you need with ohm's law.

Good luck!