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Sharp Brains, can you answer?

if one uses a test light and hits the ground side after a load it must be zero as one is one the ground side. the loads on a circuit must consume all the voltage. if they dont then after them theres another resistance.ie take a test light and put it on the ground side of your car's head light and it will dim.

With the value he gives of R1, R2 and Vin the drop after R1 is 1.33v, and the rest over R2 (ending with 0v), but he obviously as you imply has another R (almost dead short) value not given because he gives a load of 1A, as with only R values given the load would be 167 microA not 1A...
 
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With the value he gives of R1, R2 and Vin the drop after R1 is 1.33v, and the rest over R2 (ending with 0v), but he obviously as you imply has another R (almost dead short) value not given because he gives a load of 1A, as with only R values given the load would be 167 microA not 1A...
Thanks:wave. Please relate your reply or the 1A to the OP as well as posts #4 & 5
 
Sharp brains!!! Can you answer?

I went to buy a cellphone currently sold at £50. I didn't have it. I went to my mom and borrowed £25 and also £25 from my dad. Now I have £50.Luckily for me, when I went to buy the cellphone I bought it at £45. I have £5 left. I went back to my parents and gave £1 back to my mom and £1 to my dad. I am now left with £3. I still owe my parents £24 each. Mathematically, £24 + £24 = £48k + £3 (the remainder after I had given a pound back to each)= £51

I.e., £24 + £24 + £3 = £48 + £3 = £51

Now, where did the extra £1 come from since I only had £50

Prove your brain!!!!

As if it isn't bad enough you are a European football fan. Now you use pounds (or whatever that symbol is) instead of US dollars. What are we going to do with you, Classik?
 
As if it isn't bad enough you are a European football fan. Now you use pounds (or whatever that symbol is) instead of US dollars. What are we going to do with you, Classik?
Vote me in - as US president, and I will end American Football. :lol
 
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