Introduction to Electric Circuits

44 DC circuit analysis

Example 3.1

Determine the current flowing in the resistor R in the circuit of Fig. 3.6.

Solution
Let the currents I~, 12 and 13 flow in the three branches as shown. Applying KCL
to the node B we see that

I 1 = 12 --[-- %=::>13 -- I1- 12

We therefore effectively have two unknown currents and we need two

independent equations to solve for them. The first of these we obtain by

applying KVL to the closed path FABEF which gives, taking the clockwise,

direction around the path to be positive,

V1-311-212=0

Putting V~ - 6 and rearranging, we get

3I~ + 212 = 6 (3.6)

The second equation is obtained by applying KVL to the closed path EBCDE

to give

212 - 613 - V2 = 0

Putting V2 = 4, 13 = 11 - 12 and rearranging, we have

212 - 6(I1 - 12) = 4

212 - 6/1 + 612 = 4

• 611 + 812 = 4 (3.7)
  Since we wish to determine 12 it is convenient to eliminate/1. We can do this by
  multiplying Equation (3.6) by 2 and adding it to Equation (3.7). Thus
  6/1 + 412 = 12 (3.8)
  -6/1 + 812 = 4 (3.7 bis)
  12/2 = 16
  12 = 1.33A

l

I=,AI

Figure 3.7

A I2 .4~ ]

I~~+ ~'4i 2 -

D

~B

V=6V