Note that I put DC operating voltages at key points in the schematics. Using these and Ohm's law, you can see that the plate current in V1 is 1 mA, in the voltage divider is 0.36 mA, and in V2 is 2.5 mA. So, if we insert a 10 kΩ series resistor in the B+ line to the V1 stage (1.35 mA total), it will reduce B+ to that stage by about 13.5 V, which is inconsequential. Adding a bypass capacitor at the V1 side will reduce existing ripple by approximately the ratio of 10 kΩ to the capacitive reactance of the C. At 120 Hz (full-wave ripple of 60 Hz supply), the reactance of 20 µF is 67 Ω. This will give 43 dB ripple reduction (to about 1% of its original value). For 50 Hz supply, it will be about 1.5 dB less. This filter should also have enough attenuation at sub-sonic frequencies to stop any "motor-boat" oscillation