Assuming you build a 48-volt power supply.

The telephone = 600 Ohms as previously stated so the problem is a
simple resistive divider in the off-hook condition.

        12 volts through 600 ohms = .02 or 20 milliamps.
v(t) = 48 Volts
v(R(s)) = 36 volts.
Since the current for the whole circuit = 20 milliamps, all that is
left is to solve for R(s)
36/.02 = 1800 Ohms.

        If you build the 48-volt power supply and connect a 1800-Ohm
resistor in series, you will make a circuit that looks just like the
telephone line as far as the phone is concerned.  If you don't mind a
little mains ripple, your 48-volt supply doesn't have to be much more
than approximately a 30-volt secondary connected to a bridge and filter
capacitor.  The output voltage of the un-loaded bridge with the filter
capacitor across it will be the square-root of two times the RMS
voltage of the transformer's secondary.

        A 35-volt transformer should give you about 50 volts to play
around with.  Use at least a 100uf filter to cut down on the ripple.

        The actual voltage is not terribly important as long as it is
within five or 10 volts of 48 volts.  Our Northern Telecom DMS100 that
serves Stillwater sends us about 51 volts.  The old #5 cross bar
switch we had before that sent right at 48 volts.  Other systems may
vary a bit.  Your telephone circuit should be very forgiving of line
variations because they are out there and they can be nasty.

Martin McCormick