Thank you for reporting, we will resolve it shortly
Q.
If the current through $ 3\,\Omega $ resistor is 0.8 A, then potential drop through $ 4\,\Omega $ reisistor is:
ManipalManipal 2000
Solution:
Current across the $ 3\,\Omega $ resistor $ =0.8\text{ }A $ The current through the parallel resistance of $ 6\,\Omega $ . is $ =\frac{0.8\times 3}{6}=0.4A $ Hence, the total current across Q is $ =0.8+0.4=1.2 $ Now, potential across $ 4\,\Omega $ resistor is $ =4\times 1.2=4.8 $