#### liketesla

Joined Jul 16, 2005
3
Try these steps:

First redraw the circuit and change the direction of I2 to clockwise. I find that it makes circuit analysis because it becomes more systematic.

Second, appy KVL around both loops to find a system of two equations in two unknowns. I found the following two eqs.

6*i1 - 12 + 3 + 3(i1 - i2) = 0

3(i2 - i1) - 3 + 16i2 = 0

After solving these two equations for i1 and i2, I found i2=.333 amps, and i1=1.111.

but remember that i2 must be negated because it was originally drawn in the opposite direction.

so the final answer is i1=1.11 amps, and i2 = -.333 amps

#### liketesla

Joined Jul 16, 2005
3
Whoops, forgot the rest of it.

I think most people would assume the bottom node to be ground, so that is what I did. So using Ohms law, the node between the 12v source and 6 ohm resister is
-6.66 v. That would make the top node = 5.34 v. Then if we travel down the center branch, the voltage source causes the center node to be reduced by 3 volts, resulting the that node voltage = 2.34 v.

Hope that makes sense.

#### ecjohnny

Joined Jul 16, 2005
142
Thanks. joejester and liketelsa.
sorry..One more question here :

A parallel-plate capacitor has a capacitance of 300pF. it has 9 plates, each
40mm x 30mm, separated by mica having a relative permittive of 5. Calculate the thinkness of the mica.

The answer is given but i just cant get the working right. Can someone nice work this out for me???

#### ecjohnny

Joined Jul 16, 2005
142
its in my book. The answer in the book might be wrong anyway.

#### liketesla

Joined Jul 16, 2005
3
Which book are you working from and what is the problem number?