Numerical application of current = charge / time If 40 coulombs of charge pass a point in 20 seconds, what is the current flowing at that instant?

Introduction / Context:Current measures the rate at which electric charge flows. Simple problems using I = Q / t reinforce dimensional thinking and the relationship between charge (coulombs), time (seconds), and current (amperes).

Given Data / Assumptions:

• Total charge passed, Q = 40 C.
• Time interval, t = 20 s.
• Assume steady average over the interval.

Concept / Approach:Use the fundamental definition of average current: I = Q / t. Ensure units are SI to avoid conversion errors. A coulomb per second equals one ampere by definition.

Step-by-Step Solution:

Verification / Alternative check:Dimensional check: coulomb per second is ampere, so the units are consistent. Sanity check: halving time at fixed charge doubles current, which matches the numbers here (40/20).

Why Other Options Are Wrong:

• 0.5 A: would correspond to 10 C in 20 s, not 40 C.
• 20 A or 40 A: orders of magnitude too high; would require 400 C or 800 C in 20 s.

Common Pitfalls:Confusing Q/t with t/Q, or mixing milliseconds and seconds. Always keep SI units unless stated otherwise.

Final Answer:2 A