## Battery Charger (wattage)

1. The problem statement, all variables and given/known data

Determine the variable resistance R if the 25 Watts to be absorbed by the battery (0.035 ohm resistor and the 10.5 V element).

2. Relevant equations
V = IR
P = IV

3. The attempt at a solution

Well I did KVL clockwise current flow
-13 + 0.02*i + iR + 0.035*i + 10.5 = 0
2.5 = 0.02i + 0.035i + iR

Then factoring in 25W=10.5*i
i = 2.381 A which is close to the answer but doesn't factor in 0.035 and I'm not sure how to do it
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 Quote by jesuslovesu 3. The attempt at a solution Well I did KVL clockwise current flow -13 + 0.02*i + iR + 0.035*i + 10.5 = 0 2.5 = 0.02i + 0.035i + iR Then factoring in 25W=10.5*i i = 2.381 A which is close to the answer but doesn't factor in 0.035 and I'm not sure how to do it
The question was to compute R such that you got 25 W dissipated in the 10.5V battery. You computed what the current through this battery must be. Now use that in the equation with i and R you wrote above