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True, one could use a single series resistor of 100k to form a potential divider with the 47k input impedance of the amplifier, to give an attenuation of 10dB, however that would make the effective source impedance > 100k.

It is preferable to include a shunt resistor in the network.

Out of curiosity, I measured the component values for some commercial in-line attenuators/attenuating cables.

The first was a modified cable to allow a CD player to be used into the ‘Radio’ input of a Quad 33 preamp. It provides 14dB attenuation and uses an 8.2k series resistor and a 2k shunt resistor. The impedance of the ‘Radio’ input of the Quad 33 is 100k, so the attenuation factor is 14.3dB. The source will ‘see’ a load of 10.16k, that is ~ 100x that of the source impedance. The preamp will see a source impedance of 1.61k.

The second example was with a switched in-line attenuator, offering 10, 15 or 20 dB attenuation. For all settings the series resistor was measured to be 10.13k (i.e. a standard 10k resistor) with measured shunt resistor values of 4.68k, 2.17k and 0.99k respectively (i.e. 4.7k, 2.2k and 1k standard values). Assuming the load impedance is 47k, and then the respective attenuation values are 10.64dB, 15.46dB and 21.25dB.

The source impedance as seen by the amplifier will be 3k, 1.72k and 0.89k respectively. The load impedance as seen by the source will be 14.36k, 12.2k and 11.1k respectively.

In all of the above I have assumed a source output impedance of 100 Ohm.

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