Introduction to Electric Circuits

54 DC circuit analysis

Figure 3.26

[~R1

~ V2

R3

t i A

By current division

IL2 = [R~/(R~ + R3)]I2 = [(5/15) X (10/13.33)] = 0.25 A

Now in Equation (3.12),

Isc = ILx + IL2 = 1 + 0.25 = 1.25 A

To determine Rsc (in Equation (3.12)) we remove r and replace the batteries by

short circuits to give the circuit of Fig. 3.27 in which Rsc then equals RAB:

R3

-4 1 o A

~R1 ~']R2

-oB

Figure 3.27

RAB = Rsc = R3 + R1R2,/( R, + R2)= 10 + 50/15 = 13.33 f~

The current through the load resistor in Fig. 3.23 is given by

Ic- [Rsc/(Rsc + r)]Isc = [13.33 X 1.25/(13.33 + r)] a

According to Norton's theorem this is the current which would flow through r

in the circuit of Fig. 3.22.

When r = 10 f~, IL = (13.33/23.33)1.25 = 0.71 A. The corresponding voltage

across the load is given by VL = 0.71 • 10 = 7.1 V.

When r = 100 ~, Ic -- (13.33/113.33)1.25 -- 0.15 A. The corresponding load

voltage is then given by VL = 0.15 • 100 = 15 V.

Note that using this method we need to calculate Isc and Rsc only once and,

as r varies, simply put its new value into the equation to calculate Ic and VL.

3.8 THE MAXIMUM POWER TRANSFER THEOREM

Fig. 3.28 shows the Thevenin equivalent circuit of a network. The power in the