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

$\frac{1}{R} = \frac{1}{3} + \frac{1}{2}$

$R = 1.2 ohm$

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

$\frac{1}{R} = \frac{1}{0.5} + \frac{1}{2}$

$R = 0.4 ohm$

now this whole is in series with 1.6 ohm

$R = 0.4 + 1.6 = 2 ohm$

now we can use ohm's law

$V = i R$

$8 = i \times 2$

now we have current in the circuit as

$i = 4 A$