1. **State the problem:** Find the current through the 8 ohm resistor in a circuit containing resistors of 7 ohms, 12 ohms, and 8 ohms, and two batteries of 20V and 15V, using Kirchhoff's Voltage Law (KVL). 2. **Set up the circuit loop and assume loop current:** Assume a single loop with current $I$ flowing clockwise through all resistors. 3. **Write the KVL equation:** Sum of voltage rises and drops around the loop equals zero. This includes battery voltages (+20V and -15V) and voltage drops across resistors using Ohm's law: $V=IR$. $$20V - I \times 7 \Omega - I \times 12 \Omega - I \times 8 \Omega - 15V = 0$$ 4. **Simplify the equation:** Combine resistances: $7 + 12 + 8 = 27 \Omega$. $$20 - 27I - 15 = 0$$ $$5 = 27I$$ 5. **Solve for current $I$:** $$I = \frac{5}{27}$$ 6. **Calculate numerical value:** $$I \approx 0.1852\text{ A}$$ **Final Answer:** The current through the 8 ohm resistor is approximately **0.185 A**.