Article : Andy Collinson
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DC Analysis Examples

Example 1

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 ohm

A quick check can be performed using a simulator program. Alternatively you could use Kirchoff's voltage law on the left hand loop.