Physics question below?

Question

A 58.0 ohm resistor is connected in parallel with a 124.0 ohm resistor. This parallel group is connected in series with a 19.0 ohm resistor. The total combination is connected across a 15.0 V battery.

a) Find the current in the 124.0 ohm resistor.

b) Find the power dissipated in the 124.0 ohm resistor.

Answer 1

R1=124
R2= 58
R3= 19
Rt=( R1||R2) +R3 =R1R2/(R1+R2) +R3
Rt=(124 x 58)/(124 + 58) + 19=58.5 ohms
It=Vcc/Rt = 15/58.5=0.26A
V(R1)= Vc- (It R3)=15-0.26 (19)=10.13 V

I(R1)=V(R1)/R1 = 10.13/124.0= -0.082 A

P(R1) =V^2/R= or
P(R1) = V(R1) I(R1)= 0.83 W

Answer 2

resistance of parallel resistors
1/r=1/58+1/124
r=(58*124)/(58+124)
=(7192)/(182)
=39.5165
total resistance=39.52+19
=58.5165
total current=15/58.5165
0.2563
voltage across 19 ohms
v1=0.2563*19
=4.8704
voltage across 124 ohms is
v2=15-4.8704
=10.1296
therefore
a)i=10.1296/124
=0.08169 amp
b)w=10.1296*0.08169
=0.8275 watt