resistance is futile

[mpc4000]

can anyone help me with this query

any help would be much aprreciated apologies if i've posted into to wrong section

a 6 volt car radio takes 1.5 amp. It is required to operate the radio from a 12volt supply. What must the value of the series resistance inserted into the supply lead

effectiveness resistance of radio:
R= V/I = 6/1.5 = 4 Ohm

totally ok with that part its the next part thats got me stuffed if anyone can expalin the next in idiots speak that would be much appreciated

To keep the same current of 1.5 the total resistance must be:
R(total)= R(radio) + R (Added Resistance)

Therefore R(added) = R (total ) - R(radio)

Value of the series resistance = 8 -4 = 4 Ohms

just want to know where has the 8 come from ???

Thanks in advance for your help

 

[MarcT]

The 8 is the resistance value @ 12V

 

[Behemoth]

We have here 6V on radio and there is 12 V supply. So we need series resistance which will give 6V drop.
Once you know the current I = 1.5A
R=U/I=4 Ohms (U - required voltage drop on resistor, R - resistance of the resistor)


Different approach

We need 1.5A through the radio... So whatever supply we have (U) we need series resistor which will give total resistance (R - radio + resistor) R=U/I=

in this case

=12/1.5=8

 

[mpc4000]

We have here 6V on radio and there is 12 V supply. So we need series resistance which will give 6V drop.
Once you know the current I = 1.5A
R=U/I=4 Ohms (U - required voltage drop on resistor, R - resistance of the resistor)


Different approach

We need 1.5A through the radio... So whatever supply we have (U) we need series resistor which will give total resistance (R - radio + resistor) R=U/I=

in this case

=12/1.5=8

( CLUNK) thats the sound of the penny dropping of course it is what a plonkers thanks for the speedy response and the clarification from both you guys