In the below circuit diagram, calculate:

(a) 12Ω and 8Ω are in parallel.

$\frac{1}{\text{R}}$ = $\frac{1}{12}$ + $\frac{1}{8}$ = $\frac{2+3}{24}$ = $\frac{5}{24}$

So the equivalent resistance = $\frac{24}{5}$ = 4.8Ω

2Ω is in series with this combination.

Hence the total resistance = 2 + 4.8 = 6.8Ω

(b) Potential difference across the 12Ω resistor = potential difference across 8Ω resistor

Or, 12 × 0.4 = 8 × I₂

∴ I₂ = $\frac{12 × 0.4 }{8}$ = = 0.6 A

(c) Current I = I₁ + I₂ = 0.4 + 0.6 = 1 A