boylistad

S
∆V L
120 V
100 V
0
V L
voltage across the internal resistance is V int � I FLR int � (10 A)(2 �) �
20 V, and the terminal voltage has dropped to 100 V—a significant difference
when you can ideally expect a 120-V generator to provide the
full 120 V if you stay below the listed full-load current. Eventually, if
the load current were permitted to increase without limit, the voltage
across the internal resistance would equal the supply voltage, and the
terminal voltage would be zero. The larger the internal resistance, the
steeper is the slope of the characteristics of Fig. 5.53. In fact, for any
chosen interval of voltage or current, the magnitude of the internal
resistance is given by
INTERNAL RESISTANCE OF VOLTAGE SOURCES ⏐⏐⏐ 147
(5.15)
For the chosen interval of 5–7 A (DI L � 2A)onFig. 5.53, DV L is
4V,and R int � DV L ⁄DI L � 4V⁄ 2A� 2 �.
A direct consequence of the loss in output voltage is a loss in power
delivered to the load. Multiplying both sides of Eq. (5.13) by the current
I L in the circuit, we obtain
I LV L � I LV NL � I 2 LR int
Power Power output Power loss in
to load by battery the form of heat
(5.16)
EXAMPLE 5.21 Before a load is applied, the terminal voltage of the
power supply of Fig. 5.54(a) is set to 40 V. When a load of 500 � is
attached, as shown in Fig. 5.54(b), the terminal voltage drops to 38.5 V.
What happened to the remainder of the no-load voltage, and what is the
internal resistance of the source?
Solution: The difference of 40 V � 38.5 V � 1.5 V now appears
across the internal resistance of the source. The load current is
38.5 V/0.5 k� �77 mA. Applying Eq. (5.14),
Rint � � VNL
40 V
� � RL ���0.5 k�
I 77 mA
L
∆I L
1 2 3 4 5 6 7 8 9 10
I FL
FIG. 5.53
V L versus I L for a dc generator with R int � 2 �.
Rint � � DVL
�
D
I L
� 519.48 ��500 ��19.48 �
I (A)
POWER SUPPLY POWER SUPPLY
–
40 V
(no load)
+
(a)
R L
(b)
FIG. 5.54
Example 5.21.
I L
+
500 � 36 V
–