Study the diagram :

Given,

V = 4 V

Let,

R₁ = 10 Ω,

R₂ = 6 Ω,

R₃ = 12 Ω and

R₁ = 4 Ω

(a) As, R₁ and R₂ are in series then

R_S = R₁ + R₂ = 10 + 6 = 16 Ω

Similarly,

R₃ and R₄ are also in series,

R_S' = R₃ + R₄ = 12 + 4 = 16 Ω

Now, R_S and R_S' are in parallel combination then

$\frac{1}{\text R_\text{eq}}=\frac{1}{\text R_\text{S}}+\frac{1}{\text R_\text{S'}} = \frac{1}{16}+\frac{1}{16}=\frac{2}{16}$

⟹ ${\text R_\text{eq}} = \frac{16}{2} = 8\ \text Ω$

So, total resistance of the circuit is 8 Ω.

(b) From Ohm's law :

$\text {Current (I)} = \frac{\text V}{\text R_\text{eq}}=\frac{4}{8}=0.5\ \text A$

Current drawn from the cell is 0.5 A.

(c) As potential difference across 10 Ω and 6 Ω resistances is same as potential difference across 12 Ω and 4 Ω resistances which is V = 4 V.

Now,

$\text {Current through 10 Ω resistance} = \frac{\text V}{\text R_\text{S}}=\frac{4}{16}=0.25\ \text A$ .....equation 1

and

$\text {Current through 12 Ω resistance} = \frac{\text V}{\text R_\text{S'}}=\frac{4}{16}=0.25\ \text A$ .....equation 2

From equations 1 and 2

Current through 10 Ω resistance = Current through 12 Ω resistance = 0.25 A

So, the current through 10 Ω resistor is equal to the current through the 12 Ω resistor.