Correct option is DThe resultant resistance when 80Ω and 120Ω connected in parallel1R=180+11201R=3+2240R=2405=48Current (i)=VR=1248=0.25A (≅0.3A)\text{The resultant resistance when } 80\Omega \ \text{and} \ 120\Omega \ \text{connected in parallel} \\\frac{1}{R} = \frac{1}{80} + \frac{1}{120} \\\frac{1}{R} = \frac{3 + 2}{240} \\R = \frac{240}{5} = 48 \\\text{Current } (i) = \frac{V}{R} = \frac{12}{48} = 0.25A \ (\cong 0.3A)The resultant resistance when 80Ω and 120Ω connected in parallelR1=801+1201R1=2403+2R=5240=48Current (i)=RV=4812=0.25A (≅0.3A)