Introduction to Electric Circuits

46 DC circuit analysis

Solution

The circuit represents a battery V 1 in parallel with a second battery V2 supplying

a load resistor RL = 10 1). The terminals A and B are called the load terminals.

The resistors R~ and R 2 represent the internal resistances of the batteries V~ and

V2, respectively. To apply the principle of superposition, we first remove the

battery V1 to give the circuit of Fig. 3.9.

D

V2 u

RL

Figure 3.9

Note that the resistor R 2 is in series with the parallel combination of R~ and

RL so that the current 11 is given by

11 = Vz/[R 2 + (RIRL/R 1 + RL) ] = 12/[10 + 150/25)1 = 12/16 = 0.75 A

Using the current division technique we see that the current through the load

resistor RL is given by

ILl-- IRi/(R1 + RL)}I 1 = 15 X 0.75/(15 + 10) = 0.45 A flowing from A to B

Next we reconnect the battery V 1 and remove the battery V2, replacing it by its

internal resistance (10 1)). This results in the circuit of Fig. 3.10 in which the

I2 X ,A

R2 RL

B

Figure 3.10

resistor R 1 is in series with the parallel combination of R 2 and R L. The current I2
is therefore given by

I2 = V1/[R1 + (RERL/(R2 + RL)] = 36/[15 + 100/20)] = 36/20 = 1.8 A

At node X this current will divide equally between the resistors R 2 and R L
because they are of equal value. Thus

/L2 -- 0.9 A flowing from A to B

The current which would flow through the load resistor RL when both