Combined Series-Parallel Circuits | CK-12 Foundation - Free Printable

- The circuit consists of a 24V voltage source and three resistors: 110Ω, 220Ω, and 180Ω.
- The 220Ω and 180Ω resistors are connected in series, forming a combined resistance of 220 + 180 = 400Ω.
- This 400Ω combination is connected in parallel with the 110Ω resistor.
- Ammeter A1 measures the total current from the source.
- Ammeter A2 measures the current through the 220Ω resistor (which is the same as through the 180Ω resistor since they are in series).
- Ammeter A3 measures the current through the 180Ω resistor (same as A2).

Step 1: Calculate equivalent resistance of the parallel branches.
- Branch 1: 110Ω
- Branch 2: 220Ω + 180Ω = 400Ω
- Equivalent resistance R_eq = (110 * 400) / (110 + 400) = 44000 / 510 ≈ 86.27Ω

Step 2: Calculate total current (A1).
- I_total = V / R_eq = 24V / 86.27Ω ≈ 0.278A

Step 3: Calculate current through the 110Ω branch.
- I_110 = V / 110 = 24 / 110 ≈ 0.218A

Step 4: Calculate current through the 400Ω branch (A2 and A3).
- I_400 = V / 400 = 24 / 400 = 0.06A
- Since 220Ω and 180Ω are in series, the current through both is the same: I_A2 = I_A3 = 0.06A

Final Answers:
- A1: 0.278A
- A2: 0.06A
- A3: 0.06A