Four-branch parallel network: With total input current of 800 mA and known branch currents of 40 mA, 70 mA, and 200 mA, what is the current in the fourth branch?
Electronics
Parallel Circuits
Difficulty: Easy
Choose an option
Answer
Correct Answer: 490 mA
Explanation
Introduction / Context:Kirchhoff's current law (KCL) states that the algebraic sum of currents entering a node equals the sum leaving. In a parallel network, total source current equals the sum of branch currents. This is a straightforward current bookkeeping problem.
Given Data / Assumptions:
- Total source current I_total = 800 mA.
- Three branch currents: 40 mA, 70 mA, and 200 mA.
- All branches are in parallel across the same source.
Concept / Approach:Let I4 be the unknown branch current. Apply KCL: I_total = I1 + I2 + I3 + I4. Solve for I4 by subtracting known currents from I_total.
Step-by-Step Solution:
Sum known branches: 40 + 70 + 200 = 310 mA.Compute I4: I4 = 800 mA - 310 mA.I4 = 490 mA.Verification / Alternative check:Add all four: 310 mA + 490 mA = 800 mA, matching the total and confirming KCL.
Why Other Options Are Wrong:
- 800 mA: That is the total, not a single branch.
- 310 mA: Sum of the given three branches, not the unknown.
- 0 A: Would imply no fourth branch contribution, which contradicts the given total.
Common Pitfalls:
- Mistakes in arithmetic subtraction with milliampere values.
Final Answer:490 mA