I'm reading the book "Practical Electronics for Inventors, 4th Ed", and I find it great, but I'm stuck on this (seemingly straightforward) section on page 321:
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First off, the values in the "Adjusting Load Current" graph are wrong.
As the author wrote, when a-to-b Resistance is 10kΩ, V-load is 5V, not 1V, and I-load is 0.5mA, not 0.1mA.
What bothers me, however, is the graph for the "Adjusting Load Voltage" circuit.
Wrong values in the graph, ok.
But how come I-load doesn't follow the same curve as V-load?
It's a DC circuit with plain resistors.
When a-to-b Resistance is 5kΩ, V-load is 4V, so I-load becomes 0.4mA (I = V/R = 4V/10kΩ = 0.4mA).
When a-to-b Resistance is 1kΩ, V-load is 8.26V, so I-load becomes 0.826mA (I = V/R = 8.26V/10kΩ = 0.826mA).
Again, it might have been an error with the graph, but the author explicitly points out the difference:
Adjusting load current:
Notice in the graph that the load current follows a curve similar to that of the voltage.
...
Adjusting load voltage:
Notice that in the graph, the load current doesn’t fall as quickly as in the previous configuration. In fact, from 0 to 5 kΩ the current falls about only 1/10 its maximum value. However, from 5 kΩ on, the drop grows significantly.
What am I missing?
