Nodal analysis find the power dissipation, Mesh analysis to find currents,?

Question

please download word document from following link, it contain diagrams. Thanks
http://siel.edu.pk/PhyQuestions.doc

Please now Q2 where Use Mesh analysis to find currents through all loops also find currents through each resistance in the networks below. Identify and label each mesh and also show each step of calculation otherwise you will lose your marks. Also mention the units of each derived value.

and last question in uploaded document is to Use Mesh analysis to find Current through each mesh in the given network also find the current Io. Identify and label each mesh and also show each step of calculation otherwise you will lose your marks. Draw the complete circuit diagram and also mention the units of each derived value.

Answer 1

Q1: Find the power through the 6k ohm R using nodal analysis



First, I will reduce the 3 and 2 k ohm resistors to one

1/R=1/3+1/2
R=1.2 kohm

summing the voltages around the right
I*1.2 kOhm+(I-2 mA)*6 kOhm=3 V
This also took into account the current at the node where the 1.2 kOhm , the 6KOhm and 2 mA current source intersect.

Since V=I*R and the current through the 1.2 Kohm is the sum of the currents through the 6KOhm and 2 mA source

Also, since power is V*I and V=I*R, Power is I^2*R

solving for I
I=2.08 mA
check
2.08333*1.2+(.08333*6)
=3V checks out

So the current through the 6kOhm resistor is 0.8333 mA

P=0.8333^2*6/1000
=4.17 mW

First, the current through R4 is
3/8 A since the voltage is 3V

The current through R1 is 6A


summing the currents at the R3, R4, 3V node
I4-I3-I3V=0
At the R1, 4V, R2 node
I2+I3-6=0

at the bottom node
I4+I3V+I2+6=0

substituting and doing algebra
2*I4-2*I3+12=0
since I4=3/8

2*I3=3/4+12
I3=6.375 A

I2=6-I3
I2=0.375 A
I3V=3/8-6.375
I3V=-6A

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Answer 2

If the circuit is linear, the currents scale immediately with resistance. You prof is in simple terms saving time via no longer many times writing kilo. as quickly as he determines the currents he will would desire to scale them down via an element of 1000.