Charge–current calculation in basic electricity: If 6 A flows through a lamp filament for 1.75 s, how many coulombs of electric charge pass through the filament?

Introduction / Context:
Relating current, charge, and time is a core skill in circuit analysis. This relationship is useful for understanding capacitor charging, pulse currents, and energy transfer over time in basic electronics and power systems.

Given Data / Assumptions:

• Current I = 6 A (amperes).
• Time t = 1.75 s (seconds).
• We are asked to find the total charge Q that flows.

Concept / Approach:
The fundamental relation is Q = I * t, where Q is in coulombs, I in amperes, and t in seconds. One ampere equals one coulomb per second, so multiplying current by time yields charge directly.

Step-by-Step Solution:

Verification / Alternative check:
Check units: A * s = C, which is dimensionally correct. A quick mental estimate: 6 * 2 = 12 C; since time is slightly less than 2 s (1.75 s), 10.5 C is reasonable and correct.

Why Other Options Are Wrong:

• 105 C and 34 C: Each results from using an incorrect time multiplier or arithmetic error.
• 3.4 C: Much too small; would correspond to a significantly shorter time or smaller current.

Common Pitfalls:

• Mistaking milliseconds or minutes for seconds. Always convert time to seconds before applying Q = I * t.
• Dropping a decimal (e.g., 1.75 vs 17.5) leading to a tenfold error.

Final Answer:
10.5 C