Find Current at a Specific Time from Q(t)
Question:
The charge in an electronic device satisfies the equation:
Q(t) = 4.16t³ + 2.91t + 2.16 / t² (in Coulombs)
What is the current at time t = 5.29 seconds?
(Assume units are in Amperes and round the final answer to 3 significant figures.)
Answer with Detailed Step-by-Step Explanation:
Step 1: Understand the relationship between charge and current
The current I(t) is the derivative of charge Q(t) with respect to time t:
I(t) = dQ/dt
Step 2: Differentiate Q(t)
Given:
Q(t) = 4.16t³ + 2.91t + 2.16t⁻²
Differentiating term-by-term:
d/dt [4.16t³] = 12.48t²
d/dt [2.91t] = 2.91
d/dt [2.16t⁻²] = -4.32t⁻³
So the expression for current becomes:
I(t) = 12.48t² + 2.91 − 4.32 / t³
Step 3: Evaluate at t = 5.29 s
I(5.29) = 12.48 × (5.29)² + 2.91 − 4.32 / (5.29)³
Now calculate each term:
- (5.29)² = 27.9841
- (5.29)³ = 148.0359
Substitute values:
I(5.29) ≈ 12.48 × 27.9841 + 2.91 − 4.32 / 148.0359
I(5.29) ≈ 349.2416 + 2.91 − 0.0292
Step 4: Final calculation
I(5.29) ≈ 352.1224 Amps
✅ The current at t = 5.29 seconds is approximately 352 A (to 3 significant figures).