DC Analysis Examples

Circuit : Andy Collinson

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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

On the right hand loop,

Vab

10

-6

I2 = 6 + I3 * 2 = 10V

= 16 - 4*I2

= 16 - 4*I2

= -4 *I2

= 1.5 amp

On the right hand loop,

Vab

10

-6

I2 = 6 + I3 * 2 = 10V

= 16 - 4*I2

= 16 - 4*I2

= -4 *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 therefore I1 = I3 - I2 = 0.5 amp

On the left hand loop,

Therefore :

Therefore :

Vab

10

R

10

R

= 12 - R * 0.5

= 12 - R * 0.5

= 2/0.5 = 4ohms

= 12 - R * 0.5

= 2/0.5 = 4ohms

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