Question

a series combination of 4 ohm and 8 on registration connected in parallel with the series combination of resistors of 12 m what will happen to the total resistance and current drawn by a circuit if the supply forward is connected across the combination​

Answer 1

Answer:

Two resistances connected in parallel their net resistance would be 4×8/(4+8) = 8/3 ohms

It is connected to 12 ohm resistor in series.

Therefore the net resistance is 8/3 + 12 = 44/3 ohms.