call the unknown current i
6 = i^2 R outside
so R = 6/i^2
total resistance = 2 + (6/i^2)
E = i *total resistance
8 = i (2 + (6/i^2) )
8 = 2 i + 6/i
2 i^2 -8i + 6 = 0
i^2 - 4 i + 3 = 0
(i-3)(i-1) = 0
i = 3 or i = 1
which of those solutions makes sense
if i = 3 then
V loss internal = 2 * 3 = 6 volts
only 2 volts external, most power used internal, 6 of our 8 volts used internally
i = 1 makes more sense
then 2 volts lost internal and 6 of our 8 volts powers our load outside.
A battery has an emf of 8V and an internal resistance 2ohms. If it is connected to an external circuit that draws 6W of power, what is a possible current in the circuit?
Please give me some ideas to do it! THANKS!
1 answer
1 answer