Originally Posted by robemcdonald
Being an electrical novice I'm not really sure what that really means.
I think he means you can't fix the problem by buying a different flasher unit.
Originally Posted by robemcdonald
Am I right to think a 9/10 ohm resistor will do it?
If the indicator is meant to be 21W, then yes. You'll probably only find a 10 ohm resistor for sale, not a 9 ohm. Make sure you connect the resistor across the bulb (in parallel) not in series with it. The resistor will dissipate 16W of power (21 - 5) if the indicators get stuck on, so make sure the resistor is rated for that much.
There are other ways of arriving at your answer:
If you need to draw 21W of power to keep your flasher happy but your bulb only draws 5W, there's 16W left that needs to be drawn by your resistor.
Power = voltage x current (P = VI).
16 = 13 x I (using 13 because the voltage might be this high with the motor running)
I = 16 / 13 = 1.23 amps
What resistance will draw 1.23A when connected to a 13V source?
V = IR
13 = 1.23 x R
R = 13 / 1.23 = 10.6 ohms
Similarly if you need to draw 8W but only have a 5W bulb:
Need to draw 3W extra with the resistor.
P = VI
3 = 13 x I
I = 3 / 13 = 0.23 amps
Find resistor needed: V = IR
13 = 0.23 x R
R = 13 / 0.23 = 56.5 ohms
This resistor would only need to be rated at a minimum of 3W.
Also am I risking a fire by putting too much through the 5 Watt bulb?
Not if the bulb is rated for 12V. A 12V 5W bulb will draw 5W of power when connected to a 12V source. This is how much power it needs, and it will run happily.
Your car's electrical system will also run happily (aside from the flasher) because it's designed to be able to supply more power (either 8W or 21W) for the original design of indicator.
The ways to risk fire would be:
1.) "force feed" the bulb more power by supplying it with a higher voltage, e.g. 24V. This would cause the bulb to dissipate more power than it's designed for, and it would go very bright then blow.
2.) Draw more power from your car's electrical system than it's designed to deliver. E.g. if the car's designed to supply one 21W bulb but you put two such bulbs in parallel, you'll be asking for 42W from your car. Ideally a fuse will blow in this situation to protect your wiring from being overloaded.
I hope your 5W bulb is very efficient and bright (like LEDs) though, if it's meant to be as visible as a 21W bulb. You don't want anyone crashing into your Sprint..
) In fact it's just like what happens when you have a bulb in the circuit fail. 
. It looks as though the maximum power resistor I can get is 10 Watt - 10 Ohm. I can also get a 7 watt 10 ohm resistor. COuld I use these in series to to gie 17 watts? and then put this in parralell to the bulb? Can you use resistors in series for this? 
nice bit of help between friends got the problem solved
, brilliant 