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

q_1 = 10 × 1 = 10 C

q_2 = [(10 + 5)/2] × 1 = 7.5 C

q_3 = 5 × 1 = 5 C

q = q_1 + q_2 + q_3

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.

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

q_1 = 10 × 1 = 10 C

q_2 = [(10 + 5)/2] × 1 = 7.5 C

q_3 = 5 × 1 = 5 C

q = q_1 + q_2 + q_3

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.

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

q_1 = 10 × 1 = 10 C

q_2 = [(10 + 5)/2] × 1 = 7.5 C

q_3 = 5 × 1 = 5 C

q = q_1 + q_2 + q_3

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.