P = IV
P = V^2/Rt
P*Rt = V^2
V = sqrt( P*Rt)
V = 12.3 volts
this is your wrong answer ...
we know the current in the circuit will split at the branch nodes so
P = IV will be lower in the branch nodes if the wattage is the same for all components then the branch node items will see less power
the largest current will flow through the series resistors only
I = V/Rt for the series resistors
and Vdrop = I*R this means the largest power dissipation would be in the highest resistance and highest current.
the 40 ohm series resistor will dissipate the largest amount of power..
P = Vdrop*I
P = Vdrop * V/Rt
we know Vdrop = V*R/Rt
P = (V*R/Rt)*V/Rt
= V^2*R/Rt^2
solving for V
V^2 = Rt^2*P/R
V = sqrt( Rt^2 *P/R)
R = 40 ohms
V = 21.88 volts
its difficult to explain but I am 90 percent sure this is correct
you could do
it another way
P = I * V
P = I * I*R
P = I ^2 *R
we know I = V/Rt
so again we get ( V^2/Rt^2 )*R
where R is the resistor you are trying to measure its power ..
notice with the current equation current is more important than resistance because it gets squared ... for power
so if you double the current the power is 4 times..
same if you half the current the power is 1/4
thats why i know the power of the 50 ohm parallel resistors are lower power dissipated..
the current in the big parallel branch circuit will be about 1/3 in each branch
Ibranch = I *Rparalleltotal/R
the question was simplified because the parellel circuits had less resistance..than all the series ones.. this is the correct answer
V = 21.88