1. **State the problem:** We have a circuit with resistors R1=220 Ω, R2=130 Ω, R3=470 Ω, R4=100 Ω, R5=270 Ω, and a voltage source of 18 V. We need to fill in the table with voltage (V), current (I), resistance (R), and power (P) for each resistor and the total circuit. 2. **Analyze the circuit:** The circuit is a combination of series and parallel resistors. R1 and R2 are in series on the left branch, R4 and R5 are in series on the right branch, and R3 is in the middle branch connecting top and bottom. 3. **Calculate equivalent resistances:** - Left branch resistance: $$R_{left} = R1 + R2 = 220 + 130 = 350\ \Omega$$ - Right branch resistance: $$R_{right} = R4 + R5 = 100 + 270 = 370\ \Omega$$ 4. **Calculate total resistance:** R3 is in parallel with the combination of left and right branches in series, so first find the parallel of R3 with the combined left and right branches in series: The two branches (left and right) are in parallel with R3 in the middle. The total resistance is: $$\frac{1}{R_{total}} = \frac{1}{R_{left} + R_{right}} + \frac{1}{R3} = \frac{1}{350 + 370} + \frac{1}{470} = \frac{1}{720} + \frac{1}{470}$$ Calculate: $$\frac{1}{R_{total}} = \frac{1}{720} + \frac{1}{470} = \frac{470 + 720}{720 \times 470} = \frac{1190}{338400}$$ So: $$R_{total} = \frac{338400}{1190} \approx 284.37\ \Omega$$ 5. **Calculate total current from the source:** Using Ohm's law: $$I_{total} = \frac{V}{R_{total}} = \frac{18}{284.37} \approx 0.0633\ A$$ 6. **Calculate voltage across the parallel branches:** Voltage across the parallel branches is the same as the source voltage, 18 V. 7. **Calculate current through R3:** Current through R3: $$I_{R3} = \frac{V}{R3} = \frac{18}{470} \approx 0.0383\ A$$ 8. **Calculate current through the combined left and right branches:** Current through combined left and right branches: $$I_{branches} = I_{total} - I_{R3} = 0.0633 - 0.0383 = 0.0250\ A$$ 9. **Calculate voltage across left and right branches:** Voltage across combined left and right branches: $$V_{branches} = I_{branches} \times (R_{left} + R_{right}) = 0.0250 \times 720 = 18\ V$$ 10. **Calculate current through left branch (R1 and R2):** Since left and right branches are in series, current splits between them inversely proportional to resistance: Total resistance of left and right branches in series is 720 Ω. Current through left branch: $$I_{left} = I_{branches} \times \frac{R_{right}}{R_{left} + R_{right}} = 0.0250 \times \frac{370}{720} \approx 0.01285\ A$$ 11. **Calculate current through right branch (R4 and R5):** $$I_{right} = I_{branches} \times \frac{R_{left}}{R_{left} + R_{right}} = 0.0250 \times \frac{350}{720} \approx 0.01215\ A$$ 12. **Calculate voltage across R1 and R2:** - $$V_{R1} = I_{left} \times R1 = 0.01285 \times 220 = 2.83\ V$$ - $$V_{R2} = I_{left} \times R2 = 0.01285 \times 130 = 1.67\ V$$ 13. **Calculate voltage across R4 and R5:** - $$V_{R4} = I_{right} \times R4 = 0.01215 \times 100 = 1.22\ V$$ - $$V_{R5} = I_{right} \times R5 = 0.01215 \times 270 = 3.28\ V$$ 14. **Calculate power for each resistor:** Power formula: $$P = I^2 \times R$$ - $$P_{R1} = (0.01285)^2 \times 220 = 0.0363\ W$$ - $$P_{R2} = (0.01285)^2 \times 130 = 0.0215\ W$$ - $$P_{R3} = (0.0383)^2 \times 470 = 0.688\ W$$ - $$P_{R4} = (0.01215)^2 \times 100 = 0.0148\ W$$ - $$P_{R5} = (0.01215)^2 \times 270 = 0.0398\ W$$ 15. **Summarize results in the table:** | Resistor | V (V) | I (A) | R (Ω) | P (W) | |---------|-------|-------|-------|-------| | R1 | 2.83 | 0.01285 | 220 | 0.0363 | | R2 | 1.67 | 0.01285 | 130 | 0.0215 | | R3 | 18 | 0.0383 | 470 | 0.688 | | R4 | 1.22 | 0.01215 | 100 | 0.0148 | | R5 | 3.28 | 0.01215 | 270 | 0.0398 | | Total | 18 | 0.0633 | 284.37| 0.800 | **Final answer:** The table is filled with the above values for voltage, current, resistance, and power for each resistor and the total circuit.