1. **State the problem:** We want to limit the current flowing through a 50 ohm resistor to 10 A when connected to a 600 V power source. We need to find how to connect an auxiliary resistor and find its resistance. 2. **Analyze the situation:** - The voltage across the combined resistors is $V = 600$ V. - The desired current is $I = 10$ A. - The resistor $R_1 = 50$ ohms. 3. **Find the total resistance needed:** Using Ohm's law $V = IR$, the total resistance $R_{total}$ should be: $$ R_{total} = \frac{V}{I} = \frac{600}{10} = 60 \ \Omega $$ 4. **Determine how to connect auxiliary resistor:** - Since adding resistance in series adds values, $R_{total} = R_1 + R_{aux}$. - Since adding resistance in parallel lowers total resistance, it cannot reach 60 ohms if $R_1=50$ ohms only. So the auxiliary resistor must be connected in series. 5. **Calculate auxiliary resistance:** $$ R_{aux} = R_{total} - R_1 = 60 - 50 = 10 \ \Omega $$ **Final answer:** Auxiliary resistor should be connected in series with resistance 10 ohms.