Introduction
A parallel circuit worksheet 2 answer key is an essential tool for teachers, tutors, and students who want to verify their understanding of how components share the same voltage while allowing independent current paths. This article walks you through the complete solution set for Worksheet 2, explains the underlying concepts, and offers tips for creating similar worksheets. By the end, you’ll not only have the answer key at your fingertips but also a deeper grasp of parallel‑circuit analysis that you can apply to future assignments and lab activities Most people skip this — try not to..
Why a Dedicated Answer Key Matters
- Immediate feedback – Students can compare their calculations with the correct results, identify mistakes, and correct misconceptions on the spot.
- Teacher efficiency – Grading becomes faster, and instructors can focus on discussing concepts rather than checking arithmetic.
- Curriculum alignment – A well‑crafted answer key ensures that the worksheet matches the learning objectives outlined in most high‑school physics standards (e.g., NGSS MS‑PS2‑3, AP Physics 1).
Overview of Worksheet 2
Worksheet 2 typically follows an introductory worksheet that covers series circuits. It introduces the following topics:
- Voltage distribution across parallel branches.
- Current division using Ohm’s Law and the current‑divider rule.
- Equivalent resistance of multiple parallel resistors.
- Power calculations for each branch and for the whole circuit.
The worksheet contains six problems ranging from simple calculations (two resistors in parallel) to more complex networks (three or four branches with mixed voltage sources). Below is the complete answer key, organized problem‑by‑problem, followed by a step‑by‑step explanation of each solution Small thing, real impact..
Answer Key – Problem 1
Problem statement (summarized):
Two resistors, R₁ = 120 Ω and R₂ = 180 Ω, are connected in parallel across a 12 V battery. Find the total current supplied by the battery, the current through each resistor, and the equivalent resistance Not complicated — just consistent..
Solution
- Equivalent resistance
[ R_{\text{eq}} = \frac{1}{\displaystyle\frac{1}{R_1}+\frac{1}{R_2}} = \frac{1}{\frac{1}{120}+\frac{1}{180}} = \frac{1}{0.00833+0.00556} = \frac{1}{0 Worth keeping that in mind. Turns out it matters..
- Total current (Iₜ)
[ I_{\text{total}} = \frac{V}{R_{\text{eq}}}= \frac{12;\text{V}}{72;\Omega}=0.167;\text{A}=167;\text{mA} ]
- Branch currents (Ohm’s Law for each resistor)
[ I_1 = \frac{V}{R_1}= \frac{12}{120}=0.100;\text{A}=100;\text{mA} ]
[ I_2 = \frac{V}{R_2}= \frac{12}{180}=0.067;\text{A}=67;\text{mA} ]
- Check: (I_1+I_2 = 0.100+0.067 = 0.167;\text{A}) – matches the total current.
Answer summary:
- (R_{\text{eq}} = 72;\Omega)
- (I_{\text{total}} = 167;\text{mA})
- (I_1 = 100;\text{mA}) (through 120 Ω)
- (I_2 = 67;\text{mA}) (through 180 Ω)
Answer Key – Problem 2
Problem statement:
Three resistors (R₁ = 220 Ω, R₂ = 330 Ω, R₃ = 470 Ω) are connected in parallel across a 9 V source. Determine the equivalent resistance, total current, and the power dissipated by each resistor Worth keeping that in mind. Took long enough..
Solution
- Equivalent resistance
[ \frac{1}{R_{\text{eq}}}= \frac{1}{220}+\frac{1}{330}+\frac{1}{470} ]
Calculate each term:
[ \frac{1}{220}=0.004545,\quad \frac{1}{330}=0.003030,\quad \frac{1}{470}=0.002128 ]
[ \frac{1}{R_{\text{eq}}}=0.009703 ;\Rightarrow; R_{\text{eq}} \approx 103;\Omega ]
- Total current
[ I_{\text{total}} = \frac{9;\text{V}}{103;\Omega}=0.0874;\text{A}=87.4;\text{mA} ]
- Branch currents
[ I_1 = \frac{9}{220}=0.0409;\text{A}=40.9;\text{mA} ]
[ I_2 = \frac{9}{330}=0.0273;\text{A}=27.3;\text{mA} ]
[ I_3 = \frac{9}{470}=0.0191;\text{A}=19.1;\text{mA} ]
- Power in each resistor ((P = VI = I^2R))
[ P_1 = I_1^2 R_1 = (0.0409)^2 \times 220 \approx 0.369;\text{W} ]
[ P_2 = (0.0273)^2 \times 330 \approx 0.246;\text{W} ]
[ P_3 = (0.0191)^2 \times 470 \approx 0.171;\text{W} ]
Answer summary:
- (R_{\text{eq}} \approx 103;\Omega)
- (I_{\text{total}} = 87.4;\text{mA})
- (I_1 = 40.9;\text{mA},; I_2 = 27.3;\text{mA},; I_3 = 19.1;\text{mA})
- Powers: (P_1 = 0.37;\text{W},; P_2 = 0.25;\text{W},; P_3 = 0.17;\text{W})
Answer Key – Problem 3
Problem statement:
A circuit contains a 15 V source feeding two parallel branches. Branch A has a 100 Ω resistor, while Branch B contains two resistors (150 Ω and 300 Ω) in series. Find the voltage across each branch, the current in each branch, and the total power supplied by the source No workaround needed..
Solution
- Resistance of Branch B
[ R_{\text{B}} = 150;\Omega + 300;\Omega = 450;\Omega ]
- Branch voltages – In a parallel network, the voltage across each branch equals the source voltage:
[ V_{\text{A}} = V_{\text{B}} = 15;\text{V} ]
- Branch currents
[ I_{\text{A}} = \frac{V}{R_{\text{A}}}= \frac{15}{100}=0.150;\text{A} ]
[ I_{\text{B}} = \frac{15}{450}=0.0333;\text{A} ]
- Total current
[ I_{\text{total}} = I_{\text{A}} + I_{\text{B}} = 0.150 + 0.0333 = 0 Simple as that..
- Total power ((P = VI))
[ P_{\text{total}} = 15;\text{V} \times 0.1833;\text{A} \approx 2.75;\text{W} ]
Answer summary:
- Voltage across each branch = 15 V
- (I_{\text{A}} = 150;\text{mA}) ; (I_{\text{B}} = 33.3;\text{mA})
- (I_{\text{total}} = 183;\text{mA})
- (P_{\text{total}} \approx 2.75;\text{W})
Answer Key – Problem 4
Problem statement:
Four resistors (R₁ = 68 Ω, R₂ = 82 Ω, R₃ = 120 Ω, R₄ = 150 Ω) are connected in parallel across a 24 V supply. Determine the equivalent resistance, total current, and the percentage of total current that flows through each resistor Surprisingly effective..
Solution
- Calculate reciprocal sum
[ \frac{1}{R_{\text{eq}}}= \frac{1}{68}+\frac{1}{82}+\frac{1}{120}+\frac{1}{150} ]
Values:
[ \frac{1}{68}=0.01471,; \frac{1}{82}=0.01220,; \frac{1}{120}=0.00833,; \frac{1}{150}=0.00667 ]
[ \frac{1}{R_{\text{eq}}}=0.04191 ;\Rightarrow; R_{\text{eq}} \approx 23.9;\Omega ]
- Total current
[ I_{\text{total}} = \frac{24}{23.9}=1.004;\text{A}\approx 1.00;\text{A} ]
- Individual branch currents
[ I_1 = \frac{24}{68}=0.353;\text{A} ]
[ I_2 = \frac{24}{82}=0.293;\text{A} ]
[ I_3 = \frac{24}{120}=0.200;\text{A} ]
[ I_4 = \frac{24}{150}=0.160;\text{A} ]
- Current percentages
[ %I_1 = \frac{0.353}{1.004}\times100 \approx 35.2% ]
[ %I_2 = \frac{0.293}{1.004}\times100 \approx 29.2% ]
[ %I_3 = \frac{0.200}{1.004}\times100 \approx 19.9% ]
[ %I_4 = \frac{0.160}{1.004}\times100 \approx 15.9% ]
Answer summary:
- (R_{\text{eq}} \approx 23.9;\Omega)
- (I_{\text{total}} \approx 1.00;\text{A})
- Branch currents: 0.353 A, 0.293 A, 0.200 A, 0.160 A
- Current distribution: 35.2 %, 29.2 %, 19.9 %, 15.9 %
Answer Key – Problem 5
Problem statement:
A parallel circuit contains a 6 V battery and three branches:
- Branch 1: a single 30 Ω resistor.
- Branch 2: a 60 Ω resistor in series with a 12 Ω resistor.
- Branch 3: a 45 Ω resistor.
Find the voltage across each resistor, the total power delivered by the battery, and the energy consumed after 5 minutes of operation And that's really what it comes down to..
Solution
- Branch resistances
[ R_{\text{B1}} = 30;\Omega ]
[ R_{\text{B2}} = 60;\Omega + 12;\Omega = 72;\Omega ]
[ R_{\text{B3}} = 45;\Omega ]
-
Voltage across each branch – Same as source voltage: 6 V.
-
Branch currents
[ I_1 = \frac{6}{30}=0.200;\text{A} ]
[ I_2 = \frac{6}{72}=0.0833;\text{A} ]
[ I_3 = \frac{6}{45}=0.133;\text{A} ]
- Total current
[ I_{\text{total}} = 0.200+0.0833+0.133 = 0.4163;\text{A} ]
- Total power
[ P_{\text{total}} = V \times I_{\text{total}} = 6 \times 0.4163 \approx 2.50;\text{W} ]
- Energy for 5 minutes ( (E = P \times t) , (t = 5;\text{min}=300;\text{s}) )
[ E = 2.50;\text{W} \times 300;\text{s}=750;\text{J} ]
Answer summary:
- Voltage across each resistor = 6 V
- Currents: 0.200 A (30 Ω), 0.083 A (72 Ω), 0.133 A (45 Ω)
- Total power = 2.5 W
- Energy after 5 min = 750 J
Answer Key – Problem 6
Problem statement:
A more advanced network: a 12 V source feeds two parallel groups Surprisingly effective..
- Group A: two resistors (R₁ = 100 Ω, R₂ = 200 Ω) in series.
- Group B: three resistors (R₃ = 150 Ω, R₄ = 300 Ω, R₅ = 600 Ω) in parallel.
Calculate the overall equivalent resistance, the current supplied by the source, and the voltage drop across each individual resistor.
Solution
- Resistance of Group A (series)
[ R_{\text{A}} = 100 + 200 = 300;\Omega ]
- Resistance of Group B (parallel)
[ \frac{1}{R_{\text{B}}}= \frac{1}{150}+\frac{1}{300}+\frac{1}{600} ]
[ \frac{1}{150}=0.006667,; \frac{1}{300}=0.003333,; \frac{1}{600}=0.001667 ]
[ \frac{1}{R_{\text{B}}}=0.011667 ;\Rightarrow; R_{\text{B}} \approx 85.7;\Omega ]
- Overall equivalent resistance (A and B are in parallel)
[ \frac{1}{R_{\text{eq}}}= \frac{1}{R_{\text{A}}}+\frac{1}{R_{\text{B}}}= \frac{1}{300}+ \frac{1}{85.7} ]
[ \frac{1}{300}=0.00333,; \frac{1}{85.7}=0.01167 ]
[ \frac{1}{R_{\text{eq}}}=0.0150 ;\Rightarrow; R_{\text{eq}} \approx 66.7;\Omega ]
- Total current from the source
[ I_{\text{total}} = \frac{12}{66.7}=0.1799;\text{A}\approx 180;\text{mA} ]
- Current division between groups
[ I_{\text{A}} = \frac{V}{R_{\text{A}}}= \frac{12}{300}=0.040;\text{A} ]
[ I_{\text{B}} = \frac{V}{R_{\text{B}}}= \frac{12}{85.7}=0.140;\text{A} ]
(Checks: (0.040+0.140 = 0.180;\text{A}))
- Voltage drop across each resistor
Group A (series): same current flows through both resistors, so
[ V_{R1}= I_{\text{A}} \times 100 = 0.040 \times 100 = 4.0;\text{V} ]
[ V_{R2}= I_{\text{A}} \times 200 = 0.040 \times 200 = 8.0;\text{V} ]
Group B (parallel): each resistor experiences the full source voltage (12 V) because they share the same nodes.
[ V_{R3}=V_{R4}=V_{R5}=12;\text{V} ]
Answer summary:
- (R_{\text{eq}} \approx 66.7;\Omega)
- (I_{\text{total}} \approx 180;\text{mA})
- Group A current = 40 mA; Group B current = 140 mA
- Voltage drops: (V_{R1}=4;\text{V},; V_{R2}=8;\text{V},; V_{R3}=V_{R4}=V_{R5}=12;\text{V})
How to Use This Answer Key Effectively
- Self‑Check Routine – After solving a problem, compare each step with the corresponding solution above. If a discrepancy appears, revisit the formula you used (Ohm’s Law, series‑parallel rules).
- Error‑Tracing Tips – Most mistakes stem from forgetting that voltage is common across parallel branches while current splits. Highlight the voltage line in your sketch to keep this rule visible.
- Extension Activities –
- Change one resistor value and recompute the whole network to see how sensitive the total current is.
- Add a second battery in series with the original source and redo the calculations; this reinforces the distinction between series voltage addition and parallel voltage constancy.
Frequently Asked Questions
Q1: Why does the equivalent resistance of parallel resistors always become smaller than the smallest individual resistor?
A: Because the reciprocal sum (\displaystyle \frac{1}{R_{\text{eq}}}= \sum \frac{1}{R_i}) adds positive numbers, making (\frac{1}{R_{\text{eq}}}) larger than any single (\frac{1}{R_i}). So naturally, (R_{\text{eq}}) must be smaller than each (R_i) Small thing, real impact. Simple as that..
Q2: Can a parallel circuit have different voltages across its branches?
A: In an ideal parallel network with ideal wires, all branches share the same voltage equal to the source voltage. Real‑world factors (wire resistance, faulty connections) can cause slight variations, but they are usually negligible for classroom problems.
Q3: How do I remember the current‑divider rule?
A: The current through a branch is inversely proportional to its resistance:
[ I_k = I_{\text{total}} \times \frac{R_{\text{total}}}{R_k} ]
A quick mnemonic is “more resistance, less current.”
Q4: What if a parallel circuit includes a capacitor or inductor?
A: The analysis then involves impedance (Z) instead of pure resistance. The same series‑parallel combination rules apply, but you must use complex numbers and consider frequency‑dependent behavior The details matter here. Turns out it matters..
Q5: Is it okay to round intermediate results?
A: For classroom worksheets, keep three‑significant‑figure accuracy throughout the calculation, then round the final answer to the appropriate number of significant figures (usually 2 or 3, depending on the given data) Simple, but easy to overlook..
Conclusion
The parallel circuit worksheet 2 answer key presented here covers every problem commonly found in a high‑school physics unit on parallel networks. Here's the thing — mastery of these fundamentals prepares learners for more advanced topics such as Kirchhoff’s Laws, AC analysis, and real‑world circuit design. By following the step‑by‑step solutions, students reinforce the core concepts of voltage constancy, current division, and equivalent resistance. Teachers can use the answer key not only for grading but also as a springboard for deeper inquiry—altering resistor values, adding sources, or introducing reactive components. Keep this key handy, practice the variations, and the parallel circuit will become a familiar, intuitive part of your electrical toolbox.
