7 ... So ...0.8A leaves the battery ...0.2 A goes through the branch with the diode , so the current going through the 18 Ohm must be 0.6 A ...use V=IR to find the voltage across the 18 Ohm ....
8. ..The answer to part 1 is also the voltage across the other branch with 46 Ohm in series with the diode ...you have the current , 0.2 , and the voltage ... this will give you the total resistance in this branch (use V=IR)
9 ... now you have the resistance of the diode , and the current going through it ..
Calculate the power using P=I xI x R ....(I squared R)
