1. **State the problem:** Given a circuit with total resistance $$R_T = 4\ \Omega$$, currents and resistances $$I_1 = 4\ A$$, $$R_1 = 10\ \Omega$$, $$R_2 = 20\ \Omega$$, and unknown $$R_3$$, find: a. $$R_3$$ b. Voltage source $$E$$ c. Source current $$I_s$$ d. Current $$I_2$$ through $$R_2$$ e. Power $$P_2$$ dissipated in $$R_2$$ 2. **Find $$R_3$$:** Since resistors $$R_1$$, $$R_2$$, and $$R_3$$ are in parallel and combined they give total resistance $$R_T$$, $$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$ Plug in values: $$\frac{1}{4} = \frac{1}{10} + \frac{1}{20} + \frac{1}{R_3}$$ Calculate the sum of known reciprocals: $$\frac{1}{10} + \frac{1}{20} = \frac{2}{20} + \frac{1}{20} = \frac{3}{20} = 0.15$$ So, $$\frac{1}{4} = 0.25 = 0.15 + \frac{1}{R_3}$$ $$\frac{1}{R_3} = 0.25 - 0.15 = 0.1$$ Therefore, $$R_3 = \frac{1}{0.1} = 10\ \Omega$$ 3. **Calculate $$E$$:** Voltage $$E$$ across the parallel resistors is the same for each branch. Voltage across $$R_1$$ (using Ohm's law) is: $$E = I_1 \times R_1 = 4 \times 10 = 40\ V$$ 4. **Find source current $$I_s$$:** Source current $$I_s$$ is the sum of currents through all three resistors: Find $$I_2$$ and $$I_3$$ first. Current through $$R_2$$: $$I_2 = \frac{E}{R_2} = \frac{40}{20} = 2\ A$$ Current through $$R_3$$: $$I_3 = \frac{E}{R_3} = \frac{40}{10} = 4\ A$$ So, $$I_s = I_1 + I_2 + I_3 = 4 + 2 + 4 = 10\ A$$ 5. **Determine $$P_2$$:** Power dissipated in $$R_2$$ is given by $$P_2 = I_2^2 \times R_2 = 2^2 \times 20 = 4 \times 20 = 80\ W$$ **Final answers:** $$R_3 = 10\ \Omega$$ $$E = 40\ V$$ $$I_s = 10\ A$$ $$I_2 = 2\ A$$ $$P_2 = 80\ W$$