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?

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:

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