Determine the voltage drop across the 8 Ω resistor of the circuit shown in Fig

We have,

Net effective Resistance of Branch 1 and Branch 2 is 8 ohm and 12 ohms respectively as they are connected in series.

Current Division rule: According to the current division rule, the current through any branch is equal to the ratio of the total resistance in parallel to the branch to the total resistance, multiplied by the total current in the circuit.

Here,

i₁ = i_s × \(\frac{R_2}{R_1+R_2}\)

And, i₂ = i_s × \(\frac{R_1}{R_1+R_2}\)

Now,

i_s = 5 A

R₁ = 8 Ω

R₂ = 12 Ω

Using the Current division rule to find current in Branch 2:

\(I_{2}= \frac{5×8}{8+12}\)  =  2 Amp

Therefore, Voltage drops across an 8 Ω resistor:

\(V_{8} = 8×2 = 16 \ volts\)