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Combination Circuit Practice Problems: Test Your Series‑Parallel Skills

Quick, free quiz to check your skills in solving combination circuits. Instant results.

Editorial: Review CompletedCreated By: Ella CantyUpdated Aug 27, 2025
Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Paper art circuit with series and parallel resistors on golden yellow background for combination circuit practice quiz

This quiz helps you practice combination circuits so you can simplify series-parallel networks, compute totals, and verify each step with instant feedback. Build speed and accuracy before a test, then explore more with our electrical engineering quiz, basic physics quiz, and problem solving quiz too.

Two resistors 3 ohm and 7 ohm are connected in series. What is the equivalent resistance seen by the source?
2.1 ohm
4.2 ohm
10 ohm (correct: series resistances add)
7 ohm
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Two resistors 6 ohm and 3 ohm are connected in parallel. What is the equivalent resistance?
9 ohm
2 ohm (correct: 1/Req = 1/6 + 1/3, so Req = 2)
1.5 ohm
4.5 ohm
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A 12 V source feeds a series network of 2 ohm and 4 ohm. What is the current from the source?
6 A
3 A
1 A
2 A (correct: Req = 6 ohm, I = 12/6)
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A 10 V source feeds two resistors in parallel: 5 ohm and 10 ohm. What is the current through the 5 ohm branch?
3 A
1 A
2 A (correct: branch sees full 10 V, I = 10/5)
0.5 A
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In a series circuit, which statement about voltage is correct?
Voltages across series elements add to the source voltage (correct: KVL)
Voltage is the same as current
Each element has the same voltage drop
Voltage is inversely proportional to resistance
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In a parallel circuit, which statement about current is correct?
Each branch has the same current
Branch currents add to the source current (correct: KCL)
Branch currents are always equal
Current is the same as voltage
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A 24 V source feeds a 4 ohm resistor in series with a parallel pair of 6 ohm and 6 ohm. What is the source current?
4 A
6 A (correct: 6||6 = 3, total = 4+3=7 ohm, I = 24/7 ≈ 3.429 A)
2 A
3.429 A (correct: 6||6 = 3, total = 7 ohm, I = 24/7)
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A 10 V source feeds a 2 kOhm and 3 kOhm in series. What is the voltage across the 3 kOhm resistor?
6 V (correct: divider, V3k = 10*(3/(2+3)) )
4 V
5 V
3 V
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A 12 V source feeds two parallel resistors: 8 ohm and 24 ohm. What fraction of source current goes through the 8 ohm branch?
33%
25%
50%
75% (correct: currents inversely proportional to R; I8 : I24 = 24:8 = 3:1)
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A 9 V battery feeds a 1 kOhm in series with a parallel of 1 kOhm and 2 kOhm. What is the total resistance?
1.667 kOhm (correct: 1k||2k = 0.6667k, plus 1k)
1.5 kOhm
0.667 kOhm
2.0 kOhm
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A 10 V source feeds a series network of 1 ohm, 2 ohm, and 7 ohm. Power dissipated in the 2 ohm resistor is:
4 W
0.5 W
1 W
2 W (correct: I = 10/(1+2+7)=1 A, P = I^2*R = 1*2)
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You have a series string 3 kOhm and 9 kOhm across 24 V. If a 12 kOhm load is placed in parallel with the 9 kOhm, what is the new voltage across the 9 kOhm?
12 V (correct: 9k||12k = 5.143k, divider: V = 24*(5.143/(3+5.143)) ≈ 12)
6 V
18 V
24 V
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A 15 V source feeds a T-network: left branch 10 ohm, right branch 30 ohm, both meet at a 5 ohm to ground. What is the equivalent seen by the source?
7.5 ohm
12.5 ohm (correct: 10||30 = 7.5; then 7.5 + 5 = 12.5)
15 ohm
10 ohm
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Which statement about a balanced Wheatstone bridge (R1/R2 = R3/R4) is correct for measuring the center resistor current?
The bridge branch current is maximum
The bridge branch current splits equally
The bridge branch current is zero (correct: no potential difference across the bridge)
The bridge branch current equals the source current
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A 48 V source feeds a ladder: series 4 ohm, then a shunt of 12 ohm to ground, then series 4 ohm to the output. What is total resistance seen at the source?
11 ohm (correct: 12||4 = 3, then series 4 + 3 + 4)
6 ohm (correct: middle node sees 12 ohm to ground; looking left, 4 in series with (12||4) = 4 + 3 = 7; but equivalent of the whole ladder is (4 + (12||4) )? Re-evaluate: The shunt is between two 4 ohm series; total Req = 4 + (12||4) + 4 = 4 + 3 + 4 = 11 ohm)
8 ohm
12 ohm
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A 5 V source feeds 200 ohm in series with 300 ohm||300 ohm. What power is delivered to one of the 300 ohm resistors?
0.00765 W
0.0153 W (correct: compute as shown)
0.00625 W (correct: 300||300 = 150; total 200+150=350 ohm; I = 5/350=0.014286 A; V across parallel = I*150=2.1429 V; P = V^2/R = 2.1429^2/300 ≈ 0.01531 W)
0.0306 W
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A 1 kOhm resistor is placed across the output of a 10 kOhm-10 kOhm series divider from 20 V. What is the loaded output voltage?
10.000 V
1.818 V (correct: lower Thevenin R is 5 kOhm; load forms divider: Vout = 20*(1k/(5k+1k)) = 3.333 V? Re-evaluate: Divider midpoint Thevenin is 10 V with Rth = 5 kOhm; Vout = 10*(1k/(5k+1k)) = 1.667 V)
1.667 V (correct: Vth=10 V, Rth=5 kOhm, Vout=10*(1k/(5k+1k)))
5.000 V
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A bridge network cannot be simplified by pure series-parallel combinations when:
There is only one loop
The source is DC
All resistors are equal
A branch connects two non-adjacent nodes so no two resistors share exclusive series or parallel nodes (correct: requires star-delta or Kirchhoff)
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A 30 V source feeds a series 3 ohm to node A. From A, one branch is 6 ohm to ground, another is 6 ohm in series to node B, then 6 ohm to ground. What is the voltage at node A?
10.00 V
18 V (correct: shunt at A is 6 ohm in parallel with (6+6)=12 ohm → 6||12 = 4 ohm; total 3 + 4 = 7 ohm; I = 30/7; VA = I*4 = 120/7 ≈ 17.14 V)
12.00 V
17.14 V (correct: see reduction)
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Consider a network where two 2 kOhm resistors are in series and their midpoint is tied to ground through 2 kOhm. What is the equivalent resistance seen between the ends of the series chain?
3 kOhm
2 kOhm (correct: the grounded 2 k forms two equal dividers; using symmetry or delta transformation gives Req = 2 kOhm)
1 kOhm
4 kOhm
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0

Study Outcomes

  1. Understand Series-Parallel Fundamentals -

    Gain a clear grasp of how series and parallel connections differ and how they combine in complex circuits.

  2. Analyze Circuit Configurations -

    Break down combination circuits into simpler series and parallel sections for systematic examination.

  3. Calculate Total Resistance -

    Perform accurate resistance calculations on a variety of series-parallel circuit practice problems.

  4. Apply Circuit Analysis Techniques -

    Use Ohm's law and voltage-current relationships to solve combination circuit quiz questions confidently.

  5. Compare Simplification Methods -

    Evaluate different approaches to reducing complex circuits into equivalent resistances and choose the most efficient strategy.

  6. Evaluate Problem-Solving Accuracy -

    Assess your solutions against expected results and refine your process for improved precision in circuit analysis practice problems.

Cheat Sheet

  1. Series vs. Parallel Fundamentals -

    Understand that in series circuits, resistances add linearly (Rₛ = R₝ + R₂ + ...), while in parallel circuits, reciprocals add (1/Rₚ = 1/R₝ + 1/R₂ + ...). For example, two 10 Ω resistors in parallel yield Rₚ = 5 Ω. This concept from MIT OpenCourseWare is the cornerstone for tackling combination circuit practice problems.

  2. Identifying Subnetworks -

    Learn to spot series-parallel subnetworks by redrawing circuits step by step - combine simple series or parallel groups before moving deeper into the network. A common trick is to color-code branches to avoid confusion when dealing with complex layouts. Use this strategy in your combination circuit quiz to systematically reduce resistance.

  3. Equivalent Resistance Calculation -

    Apply the formula R_eq = (R₝ × R₂) / (R₝ + R₂) for two resistors in parallel, and generalize for n resistors using 1/Rₚ = Σ (1/Rᵢ). For series-parallel circuit practice problems, plugging in numeric values early can simplify algebra. HyperPhysics suggests always checking units and simplifying fractions to avoid arithmetic slip-ups.

  4. Use of Kirchhoff's Laws -

    When subnetworks aren't obvious, turn to Kirchhoff's Voltage and Current Laws to write loop and junction equations (ΣV = 0, ΣI_in = ΣI_out). This method from IEEE tutorials helps validate your resistance reduction steps and ensures accuracy in circuit analysis practice problems. Writing down equations clearly can prevent misassignment of polarities and current directions.

  5. Problem-Solving Mnemonics -

    Remember "Simplify, Solve, Substitute" as your mantra: simplify the circuit, solve for R_eq, and substitute back into the larger network. This memory aid encourages a systematic approach and builds confidence on combination circuit practice problems. University of Illinois materials affirm that a consistent routine reduces errors under exam pressure.

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