Textbook circuit theory often isolates concepts, but real-world bench work and licensing exams demand you synthesize them. Mixed networks are where most students and junior technicians lose points—and where real PCBs fail if components are underrated. This walkthrough dissects a classic mixed network, showing every algebraic step, the hidden traps, and how to translate a theoretical answer into a physical, purchasable component.
The Problem Statement: Mixed Series-Parallel Network
Tasks:
1. Calculate the total current drawn from the source (It).
2. Calculate the voltage drop across the parallel branch (Vp).
3. R2 has burned out on the physical board. Select a concrete, real-world replacement resistor (specifying exact value, wattage, tolerance, and a manufacturer part number) assuming the circuit operates in an enclosed chassis with a mandatory 2x power safety margin.
Method Selection and The Common Trap
Which theorem applies and why? We use Equivalent Resistance Reduction combined with Ohm’s Law and the Voltage Divider Rule. We do not need Mesh or Nodal analysis here because the circuit can be simplified to a single loop through successive reduction. According to All About Circuits, reducing parallel branches first, then adding series elements, is the most computationally efficient path for this topology.
The Trap: The most common error in series and parallel practice problems is the 'phantom voltage' mistake. Students correctly calculate the equivalent resistance of the parallel branch, find the total current, and then immediately apply Ohm's Law to the parallel resistors using the *source* voltage (24V) instead of the *remaining* voltage after the series resistor's drop. The series resistor 'steals' voltage before the current ever reaches the parallel node.
Step-by-Step Algebraic Solution
- Reduce the parallel branch (Rp):
The formula for two parallel resistors isRp = (R2 × R3) / (R2 + R3).
Algebraic substitution:Rp = (220 × 330) / (220 + 330)
Numerator:220 × 330 = 72,600
Denominator:220 + 330 = 550
Division:72,600 / 550 = 132 Ω.
(Alternative fractional method: 1/Rp = 1/220 + 1/330 = 3/660 + 2/660 = 5/660. Rp = 660/5 = 132 Ω.) - Calculate total equivalent resistance (Rt):
R1 is in series with Rp.Rt = R1 + Rp = 100 + 132 = 232 Ω. - Calculate total current (It):
Apply Ohm's Law to the entire circuit.It = Vs / Rt = 24V / 232Ω ≈ 0.103448 A(or 103.45 mA). - Calculate voltage drop across R1 (V_R1):
V_R1 = It × R1 = 0.103448 A × 100 Ω = 10.345 V. - Calculate voltage across the parallel branch (Vp):
Apply Kirchhoff's Voltage Law (KVL). The source voltage must equal the sum of the drops.Vp = Vs - V_R1 = 24V - 10.345V = 13.655 V.
Order of Magnitude: Total resistance is 232Ω. 24V / 200Ω would be 120mA. Our answer of 103.45 mA is slightly less than 120mA, which perfectly aligns with a slightly higher resistance.
Units: Volts divided by Ohms yields Amperes. Voltage drops sum to 24V (10.345 + 13.655 = 24.000). The physics hold.
Decision Tree: Selecting the Replacement Resistor
Now we translate the math into a physical part. R2 (the 220Ω resistor in the parallel branch) burned out. We must find its actual power dissipation to size the replacement correctly.
Calculate Power Dissipated by R2:
We know the voltage across the parallel branch is 13.655V. Since R2 is in parallel, the voltage across R2 is exactly 13.655V.P_R2 = (Vp²) / R2 = (13.655²) / 220 = 186.459 / 220 ≈ 0.847 W.
| Parameter | Condition / Rule | Evaluation | Concrete Pick |
|---|---|---|---|
| Resistance | Must match original R2 | 220Ω is a standard E24 value | 220Ω |
| Tolerance | General purpose DC network | 1% is overkill, 5% is standard | 5% |
| Calculated Power | P = V²/R | 13.655² / 220 = 0.847W | 0.847W |
| Safety Margin | 2x derating for enclosed chassis | 0.847W × 2 = 1.694W minimum | > 1.694W |
| Standard Wattage | Next standard tier up | 1W (fails), 2W (passes) | 2W |
| Final Part | Axial, flameproof metal film | Vishay PR02 2W series | PR02000202209JR500 |
The Verdict: Do not just swap in another standard 1/4W or 1/2W 220Ω resistor; it will burn up again. You must order a 220Ω, 2W, 5% axial metal film resistor (Vishay part number PR02000202209JR500 or equivalent Yageo FMP200JR-52-220R).
Independent Verification via Kirchhoff's Laws
How do you verify the answer independently on an exam or in a SPICE simulation? Use Kirchhoff’s Current Law (KCL) at the node where the circuit splits.
- Calculate the current through R3:
I_R3 = Vp / R3 = 13.655V / 330Ω ≈ 41.38 mA. - Calculate the current through R2:
I_R2 = Vp / R2 = 13.655V / 220Ω ≈ 62.07 mA. - Apply KCL: The current entering the node (It) must equal the sum of the currents leaving the node (I_R2 + I_R3).
62.07 mA + 41.38 mA = 103.45 mA.
This exactly matches our calculated total current (It = 103.45 mA). The math is verified, and the solution is bulletproof.
FAQ: Series and Parallel Practice Problems
What if the problem uses non-standard resistor values like 215Ω?
In textbook problems, use the exact number given. In physical bench work, 215Ω isn't a standard E24 value. You would synthesize it by placing a 200Ω and a 15Ω resistor in series, or use a trimmer potentiometer if precision is required. Always check the EIA standard resistor decade tables when moving from paper to breadboard.
How does a shorted parallel branch affect the series resistor?
If R3 shorts out (0Ω), the entire parallel branch becomes 0Ω. The total circuit resistance drops to just R1 (100Ω). The total current spikes to 24V / 100Ω = 240mA. The voltage across the parallel branch drops to 0V, and R1 will likely overheat and fail if it is only rated for 1/4W (since it would be dissipating P = I²R = 0.24² × 100 = 5.76W). This is why fuses are placed in series with the main feed.
Can I use the product-over-sum rule for three parallel resistors?
No. The (R1 × R2) / (R1 + R2) shortcut only works for exactly two resistors. For three or more, you must use the reciprocal formula: 1/Rt = 1/R1 + 1/R2 + 1/R3, or reduce them two at a time (find the equivalent of the first two, then use the product-over-sum rule on that result and the third resistor).
