Walking into an electrical fundamentals exam, the most common point of failure isn't a lack of theoretical knowledge—it's a breakdown in systematic reduction. When you need to figure out how to solve parallel and series circuits that are combined into mixed ladder networks, guessing the topology visually leads to catastrophic math errors. You need a rigid, repeatable algebraic process.
This walkthrough dissects a classic mixed-circuit exam problem. We will collapse the network step-by-step, show every algebraic manipulation, and finish with the independent verification methods that separate an A-grade paper from a failed one.
The Practice Problem: A Mixed Series-Parallel Ladder
Problem Statement: A 27V DC ideal voltage source is connected to a resistor ladder network. The current leaves the positive terminal and flows through $R_1$. It then reaches a node that splits into two parallel branches containing $R_2$ and $R_3$. After recombining, the current flows through $R_4$ before returning to the negative terminal of the source. Calculate the total current supplied by the source, the voltage drop across the parallel bank, and the individual branch currents through $R_2$ and $R_3$.
Method Selection: Why Reduction Over Nodal Analysis?
Before touching a calculator, you must decide which theorem applies. Because this circuit contains no bridging components (like a Wheatstone bridge) and no multiple sources requiring superposition, it is a purely reducible network. Applying Nodal or Mesh analysis here would result in unnecessary systems of linear equations. The correct method is Successive Series-Parallel Reduction followed by Ohm's Law expansion. For a deeper dive into when to abandon reduction for nodal analysis, refer to the All About Circuits guide on series-parallel topology.
Circuit Parameters and Target Variables
Here is the data-dense spec sheet for our network. Keep this table visible as we reduce the circuit.
| Component | Value | Topology Role | Target Variable |
|---|---|---|---|
| $V_{source}$ | 27.0 V DC | Ideal Source | Total Power ($P_T$) |
| $R_1$ | 10 Ω | Series (Pre-bank) | Voltage Drop ($V_1$) |
| $R_2$ | 20 Ω | Parallel Branch A | Branch Current ($I_2$) |
| $R_3$ | 30 Ω | Parallel Branch B | Branch Current ($I_3$) |
| $R_4$ | 8 Ω | Series (Post-bank) | Voltage Drop ($V_4$) |
Step-by-Step Algebraic Solution and Reduction
Never skip algebraic steps on an exam. Partial credit is awarded for correct setup, even if your calculator arithmetic fails. We will collapse the circuit from the inside out.
Step 1: Collapse the Parallel Bank ($R_2$ and $R_3$)
The innermost parallel node consists of $R_2$ and $R_3$. For exactly two resistors in parallel, the product-over-sum formula is algebraically cleaner than the reciprocal method.
Formula: $R_{23} = \frac{R_2 \times R_3}{R_2 + R_3}$
Substitution: $R_{23} = \frac{20 \times 30}{20 + 30}$
Arithmetic: $R_{23} = \frac{600}{50}$
Result: $R_{23} = 12 \, \Omega$
Step 2: Calculate Total Equivalent Resistance ($R_{eq}$)
Now that the parallel bank is reduced to a single $12 \, \Omega$ equivalent resistor, the entire circuit is a simple series loop containing $R_1$, $R_{23}$, and $R_4$.
Formula: $R_{eq} = R_1 + R_{23} + R_4$
Substitution: $R_{eq} = 10 + 12 + 8$
Result: $R_{eq} = 30 \, \Omega$
Step 3: Find Total Current ($I_T$)
Apply Ohm's Law to the entire reduced circuit using the source voltage and the total equivalent resistance.
Formula: $I_T = \frac{V_{source}}{R_{eq}}$
Substitution: $I_T = \frac{27}{30}$
Result: $I_T = 0.9 \, A$
Step 4: Expand Back to Find Branch Voltages and Currents
In a series circuit, current is constant. Therefore, $0.9 \, A$ flows through $R_1$, the entire $R_{23}$ bank, and $R_4$. We need the voltage drop across the parallel bank to find the individual branch currents.
Voltage across the bank ($V_{23}$):
$V_{23} = I_T \times R_{23} = 0.9 \, A \times 12 \, \Omega = 10.8 \, V$
Because components in parallel share the exact same voltage, both $R_2$ and $R_3$ have $10.8 \, V$ across them.
Branch Currents ($I_2$ and $I_3$):
$I_2 = \frac{V_{23}}{R_2} = \frac{10.8}{20} = 0.54 \, A$
$I_3 = \frac{V_{23}}{R_3} = \frac{10.8}{30} = 0.36 \, A$
Sanity Checks, Common Traps, and Independent Verification
Getting the math right is only half the battle. You must prove your answer makes physical sense. For more on verification techniques, the Electronics Tutorials DC Circuits section provides excellent baseline rules for network validation.
The Trap: Visual Misidentification of Series Nodes
The most common trap in this specific topology is assuming $R_3$ and $R_4$ are in series because they look visually adjacent on a standard schematic drawing. They are not. A series relationship requires that all current leaving one component enters the next with no alternative paths. Because the node between $R_3$ and $R_4$ also connects to $R_2$, $R_3$ and $R_4$ do not share the same current. Always trace the nodes, not the physical drawing.
Sanity Check: Order of Magnitude and Physical Limits
Before moving to the next exam question, run these two rapid sanity checks:
- The Smallest Resistor Rule: The equivalent resistance of any parallel bank must be strictly less than the smallest individual resistor in that bank. Our bank yielded $12 \, \Omega$. The smallest resistor in the bank was $R_2$ at $20 \, \Omega$. Since $12 < 20$, the parallel math is physically possible.
- KVL Loop Check: The sum of all voltage drops must equal the source voltage.
$V_1 = 0.9 \times 10 = 9.0 \, V$
$V_{23} = 10.8 \, V$
$V_4 = 0.9 \times 8 = 7.2 \, V$
$9.0 + 10.8 + 7.2 = 27.0 \, V$. The loop balances perfectly.
Independent Verification: Tellegen's Theorem (Power Balance)
If you have time, verify the answer independently using power conservation. The total power delivered by the source must exactly equal the sum of the power dissipated by every resistor.
Source Power: $P_{source} = V \times I_T = 27 \, V \times 0.9 \, A = 24.3 \, W$
Dissipated Power:
- $P_1 = I_T^2 \times R_1 = (0.9)^2 \times 10 = 8.100 \, W$
- $P_2 = I_2^2 \times R_2 = (0.54)^2 \times 20 = 5.832 \, W$
- $P_3 = I_3^2 \times R_3 = (0.36)^2 \times 30 = 3.888 \, W$
- $P_4 = I_T^2 \times R_4 = (0.9)^2 \times 8 = 6.480 \, W$
Sum of Dissipation: $8.100 + 5.832 + 3.888 + 6.480 = 24.300 \, W$. The power balances to the exact milliwatt. Your algebraic reduction is verified correct.
FAQ: Common Exam Pitfalls
Q: What if the parallel bank has three resistors instead of two?
A: The product-over-sum shortcut fails for three or more resistors. You must revert to the reciprocal formula: $\frac{1}{R_{eq}} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_5}$. Solve the right side as a common fraction, then invert the final result. Never forget the final inversion step.
Q: How do I handle internal resistance in the voltage source?
A: Treat the source's internal resistance ($r_{int}$) as an additional series resistor located immediately after the ideal voltage source. Simply add $r_{int}$ to your $R_{eq}$ calculation in Step 2 before calculating total current.
