A 67.6 ohms resistor is connected in parallel with a 103.8 ohms resistor. This parallel group is connected in series with a 20.8 ohms resistor. The total combination is connected across a 18.2-V battery. Find a) the current and b) the power dissipated in the 103.8 ohms resistor.

Answer 1

Given:
R1 = 67.6 Ohms.
R2 = 103.8 Ohms.
R3 = 20.8 Ohms.
E = 18.2 Volts.

a. R =( R1*R2)/(R1+R2) + R3 = (67.6*103.8)/(67.6+103.8) + 20.8 = 61.74 Ohms.
I = E/R = 18.2/61.74 = 0.295A.

b. V2 = E - I*R3 = 18.2 - 0.295*20.8 = 12.06 Volts.
P = V2^2/R2 = 12.06^2/103.8 = 1.40 W.