Yes
and no
The resulting resistance will be 30
hm: as you calculated:
R1 in paralel with R3 = 25
hm:,
25
hm: in series with R2 = 75
hm:,
75
hm: in paralel with R4 = 30
hm:.
BUT those resistors don't share the power evenly.
You must calculate the power dissipated by each one:
R4 will dissipate V^2/50
hm: =
1 W,
so
V should be less or equal than the square root of (1 W * 50
hm
=
7.07 V.
With a 7.07 V supply, the 75
hm: branch will have a current of 94 mA, which gives:
0.444 W in R2 ( = 94 mA^2 * 50
hm: ), and
0.111 W each in R1 and R3 (with half the current, 94/2 mA^2 * 50
hm: )
So, as you must limit the power dissipated in every resistor to 1W,
the "new" 30hm: resistor will be able to dissipate less than 2 W (in fact, just 1.666 W):
1 W + 0.444 W + 2 x 0.111 W =
1.666 W, or
7.07 V ^2 / 30
hm: = 50/30 W =
1.666 W