A resistor of 8 Ω is connected in parallel with a resistor of 12 Ω. The combination is then connected in series with a 4 Ω resistor and a 12 V battery of negligible internal resistance. What is the current through the 8 Ω resistor?

The parallel combination of 8 Ω and 12 Ω gives $R_p = \frac{8 \times 12}{8 + 12} = 4.8\,\Omega$. Total resistance = $4 + 4.8 = 8.8\,\Omega$, so total current = $\frac{12}{8.8} \approx 1.364\,\text{A}$. Voltage across parallel combination = $1.364 \times 4.8 \approx 6.55\,\text{V}$. Current through 8 Ω = $\frac{6.55}{8} \approx 0.82\,\text{A}$. Wait — recalculating: $\frac{12}{8+4.8}=\frac{12}{12.8}=0.9375\,\text{A}$; $V_p=0.9375\times4.8=4.5\,\text{V}$; $I_{8\Omega}=\frac{4.5}{8}=0.5625\,\text{A}\approx0.6\,\text{A}$. The current through the 8 Ω resistor is approximately 0.6 A.