1. **Problem statement:** Find the voltage $V_2$ across the resistor $R_2 = 3\ \Omega$ and the current $I_4$ through resistor $R_4 = 8\ \Omega$ in the given circuit. 2. **Analyze the circuit:** - The voltage source $E = 12$ V is connected in series with $R_1=4\ \Omega$. - After $R_1$, the circuit splits into two parallel branches: - Branch 1: $R_2 = 3\ \Omega$ with voltage $V_2$ across it. - Branch 2: $R_3 = 6\ \Omega$. - These two parallel branches recombine and then pass through $R_4=8\ \Omega$ where current $I_4$ flows. 3. **Calculate equivalent resistance of the parallel branches:** $$\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$$ Thus, $$R_p = 2\ \Omega$$ 4. **Calculate total resistance in circuit:** $$R_{total} = R_1 + R_p + R_4 = 4 + 2 + 8 = 14\ \Omega$$ 5. **Calculate total current from the source:** Using Ohm's law, $$I_{total} = \frac{E}{R_{total}} = \frac{12}{14} = \frac{6}{7} \approx 0.857\ \text{A}$$ 6. **Calculate voltage across the parallel combination ($V_p$):** $$V_p = I_{total} \times R_p = \frac{6}{7} \times 2 = \frac{12}{7} \approx 1.714\ \text{V}$$ 7. **Voltage $V_2$ across $R_2$ is the same as across the parallel branch 1, so:** $$V_2 = V_p = \frac{12}{7} \approx 1.714\ \text{V}$$ 8. **Find current through $R_4$:** Since $R_4$ is in series with the parallel network and $R_1$, current is the same as total current, $$I_4 = I_{total} = \frac{6}{7} \approx 0.857\ \text{A}$$