Basic Electricity Formula question
First, the question is somewhat ambiguous, and certainly not well written (with the double "parallel" in there).  This puts doubt on the writers understanding and competence.

However, it seems the question is asking what the overall power is when 1 k&Omega; and 80 &Omega; are connected in parallel across 120 V:



If so, there are a number of easy ways to solve this.

One solution is to solve for the effective resistance of 1 k&Omega; and 80 &Omega; in parallel.  Then solve for the power dissipated by that resistance when 120 V is applied to it.

Another solution is to solve for the power drawn by the two resistances separately, then add the results.  Note that the power dissipated by a resistor is the square of the applied voltage divided by the resistance.  In common units:

&nbsp; &nbsp; W = V2/&Omega;

Where W is the power in watts, V the applied EMF in volts, and &Omega; the resistance in ohms.

However, all this may be a red herring.  Note that the question says the "load"
draws 2.7 A.  That load is clearly not just the two resistances, since they don't together draw 2.7 A with 120 V applied.  If this 2.7 A is the total drawn at 120 V, then it doesn't matter that some of that 2.7 A is drawn by the two known resistances.

You might be tempted to multiply the 2.7 A by the 120 V to get the total power, but that might not be right either.  That is because the power factor or phase angle of the current with respect to the voltage was not specified.  If the two are completely in phase (power factor = 1), then the basic multiply yields the correct answer.  Again though, this was not specified, and non-unity power factor is very real in the real world.

Again, this is a badly written question.  You really need to ask the instructor for clarification about what is really being asked.  A diagram would certainly help.