How to use dividing network on subwoofer - Sony MDS-JB940 Mini Disc Player

Just connect both sets of wires from the subwoofers to the same speaker output terminals on the amplifier. Make sure you keep the positive and negative wires together. This is called parallel wiring. The Type E is a 4-ohm, single voice coil subwoofer, so wiring them like this will give the amplifier a 2-ohm load, and the power produced will be divided between the two subs.

http://crutchfield.custhelp.com/rnt/rnw/img/enduser/Mono_2-4OSVC.jpg

23x56x26=33488 then divide by 1728= sqare feet

Personally for a sealed box i would go with at least 1.2 cubic feet inside volume.
A 19" x 14" x 17" box will give you a volume of 1.6 cubic feet (most common).
You can do the math by H x W x D then divide that # by 1728 (for 12" speaker) to get your enclosure volume (cubic feet).
The best enclosure depends on what your looking for in bass response.

There is nothing to configure. Hubs simply spit a network connections almost like a TV cable splitter divides cable for separate rooms.

Hubs require a router that manages the network by providing IP and other network management tools.

20 amp fuse. to determine the correct fuse, take the power rating of the amp (240 watts) and divide it by 12 (for the 12 volt car power).

No the LOWER the resistance the MORE Output or Draw, If say it is 100 Watts into 4 Ohms, if you used a 2 ohm speaker it would draw 200 watts.(only if capable of delivering that amount of power) The LOWER the Resistance, the HIGHER the Current Draw. WATTS = AMPS x VOLTS. (This is not strictly true with amplifiers output though, as it is an AC Current, and a "Reactive" load, it has, Inductance, Capacitance and Resistance, all in one.and is worked out via a Formula) see below. http://en.wikipedia.org/wiki/Audio_power

http://www.hometheaterhifi.com/volume_1_1/v1n1spk.html

Now with Dual Voice Coils, Essentially, you can think of these speakers as TWO speakers in ONE enclosure. As such, They are basically TWO Speakers. Please see the website below it explains everything for you. http://www.diyaudioandvideo.com/FAQ/Wiring/
Everything is a function of Resistance and Ohms law.
Current (I) = Voltage (E) divided by Resistance(R).
Voltage = Current Multiplied by Resistance.
Resistance = Voltage divides by Current.

that's dividing network ckt. you can purchase that to electronic store.

check for individual speaker resistance. or in the dividing network for some burn fuse or coil.

it's an 8 ohm driver. It need to be VERY efficient, around 90-91db @ 1w. Measure the enclosure size LxWxH then divide by 1728. That will give you your cu. ft. internal airspace of the box. Then look for a driver that will effectively work in that space. Shouldn't be too expensive, paper cone, foam surround, 8ohm impedance. Hope this helps

If you can access the inside of the sub box by removing the speaker, you may be able to measure the 3 dimension of each side of the enclosure.  1 cubic foot is 12x12x12 inches cubed.  That is 1728 cubic inches per cubic foot.  If you measure the inside dimensions of one of the enclosure sides as 10x14x8 inches, then that is 1120.  1120 divided by 1728 is 0.648 cubic feet.  Alternatively, you could estimate the volumes of each side of the box by measure the outside, figuring out where the box divider board is, subtracting the width of the boards used to make and then calculate.  Otherwise, I do not know the enclosure that that sub comes in.  Good luck.  Let us know how it goes.