Difficulty: Medium
Correct Answer: 1 V
Explanation:
Introduction / Context:
Analyzing series–parallel circuits often reduces to two steps: first find the equivalent resistance of parallel sub-networks, then apply series rules such as the voltage divider. This problem illustrates that workflow and emphasizes that the same voltage appears across all elements of a parallel branch.
Given Data / Assumptions:
Concept / Approach:
First, compute the equivalent resistance of the parallel network: 1/R_p = 1/R2 + 1/R3 + 1/R4. Then form a simple series pair of R1 and R_p. The voltage across the parallel branch is found by the divider V_parallel = V_s * (R_p / (R1 + R_p)). The voltage across R2 equals the branch voltage because nodes are common in parallel.
Step-by-Step Solution:
Compute 1/R_p = 1/1000 + 1/2000 + 1/1000 = 0.001 + 0.0005 + 0.001 = 0.0025.So R_p = 1 / 0.0025 = 400 Ω.Total series resistance: R_total = R1 + R_p = 2000 + 400 = 2400 Ω.Voltage divider: V_parallel = 6 * (400 / 2400) = 6 * (1/6) = 1 V.Since R2 is in that parallel branch, V_R2 = 1 V.
Verification / Alternative check:
Calculate branch currents: I_total = 6 / 2400 = 2.5 mA. Current through R1 causes a drop of V_R1 = I_total * 2000 = 5 V. Remaining voltage across the parallel network is 6 − 5 = 1 V, confirming the divider result.
Why Other Options Are Wrong:
3 V or 5 V: would require different series/parallel proportions; they contradict divider math.6 V: would be correct only if R1 = 0 (no series drop), which is not the case.
Common Pitfalls:
Forgetting that all components in parallel share the same voltage, or miscomputing R_p by adding resistances instead of conductances (reciprocals). Keep careful track of units (kΩ vs Ω).
Final Answer:
1 V
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