DC Analysis Examples

Using the techniques of Ohm's law and Kirchoff's law currents I1 and I2 can be found and the value of resistor R, in the diagram above.

The voltage across points A to B = 6 + I3 * 2 = 10V

On the right hand loop,

Vab = 16 - 4*I2

10 = 16 - 4*I2

-6 = -4 *I2

I2 = 1.5 amp On the left hand loop

Vab = 12 - R*I1

The sign of the voltages is given by the polarity rule ( current flowing into a resistor develops a p.d.positive on the side where the arrow points in.) Using Kirchoff's current law for I1 :

I1 + I2 = I3 I1 = I3 - I2 = 2-1.5 = 0.5

amp On the left hand loop:

Vab = 12 - R*0.5

10 = 12 - R*0.5

R = 2/0.5 = 4 ohmA quick check can be performed using a simulator program.

Alternatively you could use Kirchoff's voltage law on the left hand loop.