This is a follow-on to my previous note about volts and amps.
In particular, the previous note covered mostly "linear circuit" analysis
at a very elementary level. In the real world, additional considerations
apply. In particular, there are safety considerations that demand attention.
The diagram in the previous note contained several idealizations. For one
thing, it assumed that in normal operation the Kirchhoff approximation
applies, so that there is no voltage drop except inside circuit "elements"
like batteries and resistors, i.e. no voltage drop in the wires. In the
real world, under some conditions, these are not good approximations.
It is possible to design a circuit that _relies_ on the output impedance R0
to provide safety under abnormal conditions, such as a short in the load
(RL -> 0). This is usually a bad design; usually it is better to rely on
a nonlinear element such as a fuse, rather than a linear source
impedance. But in general, the possibility must be considered. One should
not blithely say that a lower source impedance is always better.
In normal operation, the load impedance RL takes on the value
RL=R1. Assume the parasitic resistance of the wires R2 is much less than R1.
Part 1: Calculate the maximum value of that R0 can have so that the power
dissipated in R0 is less than 10% of the power dissipated in the load in
normal operation.
Part 2: Now consider abnormal short-circuit operation, where RL takes on
the value RL=0 (not RL=R1 any more). Suppose there is as strict limit Pmax
on the power that can be dissipated in the wire R2 under any circumstances,
because the wire runs through a cable tray with poor ventilation et
cetera. Calculate the minimum value of R0 to ensure that this limit is not
violated during short-circuit operation.
Show that lowering the source impedance R0 can only make normal operation
better, but can only make abnormal operation worse.
Show that depending on the values of Pmax and the other parameters, it may
be impossible to find a value for R0 that satisfies part 1 and part
2. Show how to resolve this dilemma by adding a circuit breaker to the system.