1. **State the problem:** We have a series circuit with three resistors $R_1$, $R_2$, and $R_3$. The voltage drops across them are $V_1$, 5V, and 3V respectively. The total voltage supply is 18V, and the total resistance is 36Ω. We need to find $V_1$, the supply current $i$, and the values of $R_1$, $R_2$, and $R_3$. 2. **Use the voltage law for series circuits:** The sum of voltage drops across resistors equals the total voltage supply: $$V_1 + 5 + 3 = 18$$ 3. **Calculate $V_1$:** $$V_1 = 18 - 5 - 3 = 10\text{ V}$$ 4. **Calculate the supply current $i$:** In series circuits, current $i$ is the same through all resistors. Using Ohm's law $V = IR$, total voltage $V_{total} = i \times R_{total}$: $$i = \frac{V_{total}}{R_{total}} = \frac{18}{36} = 0.5\text{ A}$$ 5. **Calculate each resistor value using Ohm's law:** $$R = \frac{V}{i}$$ - For $R_1$: $$R_1 = \frac{V_1}{i} = \frac{10}{0.5} = 20\Omega$$ - For $R_2$: $$R_2 = \frac{5}{0.5} = 10\Omega$$ - For $R_3$: $$R_3 = \frac{3}{0.5} = 6\Omega$$ 6. **Check total resistance:** $$R_1 + R_2 + R_3 = 20 + 10 + 6 = 36\Omega$$ This matches the given total resistance, confirming our calculations. **Final answers:** - $V_1 = 10$ V - Supply current $i = 0.5$ A - $R_1 = 20\Omega$ - $R_2 = 10\Omega$ - $R_3 = 6\Omega$