The current through an element is shown in Fig. Q1. Determine the total charge that passed through the element from t = 0 s to t = 3 s.
Fig. Q1: Current–time graph for the charge numerical.
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Answer & Explanation
Correct AnswerC. 22.5 C
Quick Explanation
Current is the rate of flow of charge, so the required charge is the signed area under the current–time graph from 0 to 3 s. The interval contains a 10 A rectangle, a 10→5 A trapezoid and a 5 A rectangle; their areas give 22.5 C.
Formula / Key Relation
i(t) = dq/dt ⇒ dq = i(t)dt
q = ∫03 i(t)dt
q = 10(1) + [(10 + 5)/2](1) + 5(1)
q = 10 + 7.5 + 5 = 22.5 C
1 A·s = 1 C
Concept summary: current–time graph, q = ∫i(t)dt, geometric-area method and common mistakes.
Detailed Explanation
Correct Answer
C — 22.5 C
The total charge transferred during the requested 0–3 s interval is 22.5 coulomb.
Concept Foundation — Why Area Under an i–t Graph Gives Charge
Electric current tells us how quickly electric charge is crossing a reference section of a conductor. Mathematically, i = dq/dt. Rearranging gives dq = i(t)dt, and integrating over a time interval gives the total transferred charge. Therefore, on a current–time graph, the charge is exactly the signed area under the curve.
This is not just a graph trick. The units prove the same idea: ampere × second = coulomb. If a portion of the current lies below the time axis, that area is negative because the charge flow has reversed relative to the chosen current direction.
Read Fig. Q1 Before Calculating
0–1 s: current is constant at 10 A → rectangle.
1–2 s: current decreases linearly from 10 A to 5 A → trapezoid.
2–3 s: current is constant at 5 A → rectangle.
After 3 s: the graph is visible, but it must not be included because the question stops at 3 s.
Step-by-Step Calculation
q1 = 10 × 1 = 10 C
q2 = [(10 + 5)/2] × 1 = 7.5 C
q3 = 5 × 1 = 5 C
q = q1 + q2 + q3
q = 10 + 7.5 + 5 = 22.5 C
Why the Trapezoid Formula Works
Between 1 s and 2 s the current changes linearly, so its average value over that 1 s interval is (10 + 5)/2 = 7.5 A. Average current × time therefore gives 7.5 C, exactly the same result as the trapezoid area.
Concept Extension
For a horizontal segment: charge = current × time.
For a straight sloping segment: use trapezoid area or average current × time.
For a curved segment: use integration if the current equation is known.
Area below the time axis contributes negative charge.
Final Check
Unit check
10 A × 1 s, 7.5 A × 1 s and 5 A × 1 s are all in coulomb, so the final unit 22.5 C is dimensionally correct.
Exam Strategy
First mark the exact time limits asked in the question. Then split only that interval into simple shapes and add their signed areas. This is faster and safer than trying to write one unnecessary piecewise integral.
Quick Trick
Mark 0–3 s first → area = rectangle + trapezoid + rectangle = 10 + 7.5 + 5 = 22.5 C.
Why Other Options Are Wrong
A — 17.5 C: misses part of the required 0–3 s area.
B — 15 C: cannot equal the sum of the 10 C, 7.5 C and 5 C contributions.
D — 27.5 C: overcounts the interval or uses graph area outside the requested range.
Common Mistake
Using the visible graph beyond t = 3 s, or treating the 1–2 s sloping segment as a rectangle instead of a trapezoid.
Exam Tip
For i–t graphs, think ‘charge = signed area’. Always respect the exact time interval before calculating.
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