Using a DC battery to power 20 V AC, 1 A circuit
Figure 1. AC input circuit showing DC connection points.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 2. Battery power circuit.

How it works:


Two series 12 V batteries will give you 24 V DC.
The fuse will protect the batteries and the amplifier should any faults occur.
D1 will prevent backfeed from the AC supply if the mains is connected.
IOD-IX4 (the bridge rectifier) will prevent the battery backfeeding into the transformer.


Run time


  How long do you suppose 1 Ah would run this circuit?


Let's look for some clues on the mains side. 


In Figure 1 we can see a 0.5 A fuse on a 120 V supply. From this we can calculate that the maximum power from mains would be \$ P = VI = 120 \cdot 0.5 = 60~W \$. 
Redoing the calculation for 24 V and rearranging to solve for the current we get \$ I = \frac {P}{V} = \frac {60}{24} = 2.5~A \$.
Run time is \$ \frac {Ah}{A} = \frac {1}{2.5} = 0.4~h \$ at full power.


And let's look for clues on the audio side. 


The amplifier needs to output an AC signal (positive and negative) to the 4 Ω speaker while running from a single-rail power supply. To do this we would expect the TR8 and TR9 junction to sit at mid-supply voltage while the amplifier is quiet. This is confirmed by the 12.3 V reference reading on the schematic. The 1000 µF, 50 V capacitor on the output blocks the DC reaching the speaker.
When the signal goes positive TR8 will conduct and pull the output high. When the signal goes negative TR9 will pull output low. Allowing for some losses in the transistors we might see about 20 V peak to peak or 10 V peak.
From Ohm's law we can calculate the current into the speaker as \$ I = \frac {V}{R} = \frac {10}{4} = 2.5~A ~peak\$. 


The two calculations are crude but agree. 0.4 h = 24 minutes. If you're playing heavy metal with the volume at '11' then that's about it. If you're playing occasional jazz fills at civilised volume levels your power demand will be much reduced.

Output power

You may be interested in the output power of the amplifier. We saw that the peak voltage was 10 V. The RMS (root-mean-square) voltage is given by 

$$ V_{RMS} = \frac {V_{PEAK}}{\sqrt {2}} = \frac {10}{\sqrt {2}} = 7~V_{RMS} $$

The maximum speaker power for an undistorted sinewave is given by

$$ P = \frac {V^2}{R} = \frac {7^2}{4} = 12~W $$