Question

What is the current flowing through the 1.6 ohm resistance in the circuit?

Answer 1

here we need to find the net equivalent resistance of whole circuit

in order to find this

3 ohm and 2 ohm in first branch in parallel

`(1)/(R) = (1)/(3) + (1)/(2)`

`R = 1.2 ohm`

now this combination of 2 ohm and 3 ohm is in series with 0.8 ohm

so this combination is equivalent as

`R = 1.2 + 0.8 = 2 ohm`

now this 2 ohm and 0.5 ohm is in parallel

so we will have

`(1)/(R) = (1)/(0.5) + (1)/(2)`

`R = 0.4 ohm`

now this whole is in series with 1.6 ohm

`R = 0.4 + 1.6 = 2 ohm`

now we can use ohm's law

`V = i R`

`8 = i * 2`

now we have current in the circuit as

`i = 4 A`