From Current to Transferred Charge
A current describes how quickly electric charge flows. If the current stays constant, the charge transferred over a time interval is the current multiplied by the length of that interval. When current varies with time, a single multiplication is not enough: the rate may be different at different moments.
Integration combines the contributions from those changing rates. Because current has a direction relative to a chosen reference direction, its sign matters. Integrating the signed current gives net charge: charge transferred in the reference direction minus charge transferred in the opposite direction.
Definition and Units
Choose a reference direction for the current. The current \(I(t)\) is positive when charge flows in that direction and negative when charge flows in the opposite direction. Conventional current uses the direction of positive-charge flow; the sign convention must be used consistently throughout a calculation.
One ampere is one coulomb per second, so \(1\text{ A}=1\text{ C/s}\). Consequently, multiplying a current by a time gives charge: \((\text{C/s})(\text{s})=\text{C}\). The same unit check applies to the integral, whose integrand is current and whose integration variable is time.
For a constant current \(I(t)=I_0\), the integral reduces to \(I_0(b-a)\). For a varying current, integration accounts for its values throughout the interval. The integral is signed: a negative value means that the net transfer is opposite to the chosen reference direction.
Why Integration Gives Charge
Current is the rate of change of charge with respect to time. If \(q(t)\) denotes the accumulated signed charge at time \(t\), then the current is \(I(t)=q'(t)\) wherever this derivative exists. The definite integral adds the changes in charge over the interval.
For a small time width \(\Delta t\), the current is approximately constant over that piece, so the charge change is approximately \(I(t_i^*)\Delta t\) for a time \(t_i^*\) in the piece.
Summing the contributions gives an approximation to the net charge. Contributions with negative current subtract from contributions with positive current.
As the widths of the pieces approach zero, the sums approach the definite integral of the current, provided the current is integrable.
In symbols, these sums have the form \(\sum_i I(t_i^*)\Delta t_i\), and their limit is \(\int_a^b I(t)\,dt\). This explains why the integral is a signed accumulation rather than simply a rate evaluated at one time.
A useful equivalent calculation follows from the Fundamental Theorem of Calculus. If \(q\) is an accumulated-charge function with \(q'(t)=I(t)\), and the usual conditions for the theorem hold (for example, \(I\) is continuous on \([a,b]\)), then
Thus the integral gives the change in accumulated charge, not necessarily the total charge present at the end. Any initial charge \(q(a)\) is accounted for by subtracting it.
Worked Calculations
Worked Example: A Linearly Increasing Current
A current is \(I(t)=2t+1\) amperes for \(0\leq t\leq 3\), where \(t\) is in seconds. Find the net charge transferred.
Integrate the current over the stated interval:
At the upper endpoint, \(3^2+3=9+3=12\). At the lower endpoint, \(0^2+0=0\). Therefore,
The current is positive throughout the interval, so the net transfer is in the reference direction. As a check, \(q(t)=t^2+t\) is an accumulated-charge function starting from \(q(0)=0\), and \(q'(t)=2t+1=I(t)\). Its change is \(q(3)-q(0)=12-0=12\text{ C}\), as calculated. The units also agree: amperes times seconds give coulombs.
Worked Example: A Continuous Current That Reverses Direction
Suppose \(I(t)=6-2t\) amperes on \(0\leq t\leq 5\), with time in seconds. Find the net charge transferred and determine its direction.
The current is zero when \(6-2t=0\), so \(t=3\) seconds. It is positive before \(t=3\) and negative after \(t=3\). For the net charge, integrate the signed current over the full interval:
The positive result means the net transfer is in the reference direction. To check how the two portions contribute, integrate separately. From \(0\) to \(3\), the contribution is
From \(3\) to \(5\), the contribution is negative:
Adding the signed contributions gives \(9+(-4)=5\text{ C}\), matching the integral over the full interval. As a further check, \(q(t)=6t-t^2\) satisfies \(q'(t)=6-2t=I(t)\), and \(q(5)-q(0)=5-0=5\text{ C}\).
Direction Changes and Continuity
When a current is continuous and has opposite signs at two times, the Intermediate Value Theorem guarantees that it is zero at some time between them. In the preceding example, continuity and the sign change are consistent with the zero at \(t=3\). At a zero of the current, the instantaneous rate of charge transfer is zero; the charge accumulated up to that time does not have to be zero.
Continuity matters. A discontinuous current can change sign without ever taking the value zero. For example, a current may be positive before a switching time and negative at and after that time. A sign reversal in that model does not imply that there is a time at which the current equals zero. The integral can still be calculated if the current is integrable, including when it is piecewise continuous.
When a current is defined by different formulas on subintervals, integrate each formula over its own interval and add the signed results. A value assigned at a single switching instant does not change the definite integral; changing a function at finitely many individual points does not affect its integral.
Worked Example: A Discontinuous Direction Reversal
Let the current be \(1\) ampere for \(0\leq t<2\) seconds and \(-2\) amperes for \(2\leq t\leq 4\) seconds. Find the net charge over \(0\leq t\leq 4\).
Integrate on the two constant-current intervals:
The negative answer means the net transfer is opposite to the reference direction. The current is \(1\) ampere before \(t=2\) and \(-2\) amperes from \(t=2\) onward; it never equals zero in this model. Thus the direction changes discontinuously rather than by passing through a zero current.
For a check, the accumulated charge relative to \(q(0)=0\) is \(q(t)=t\) for \(0\leq t\leq 2\), and \(q(t)=2-2(t-2)\) for \(2\leq t\leq 4\). These expressions agree at \(t=2\), where both give \(2\text{ C}\). At the endpoint, \(q(4)=2-2(2)=-2\text{ C}\), confirming the integral.
A Common Sign Mistake
A common error is to treat every interval's contribution as positive, even when the current is negative. In the piecewise example, an incorrect calculation would be
This replaces the negative current \(-2\text{ A}\) by its positive magnitude and therefore does not calculate net charge. The correct signed contribution on the second interval is \((-2\text{ A})(2\text{ s})=-4\text{ C}\), so the net charge is \(2-4=-2\text{ C}\). A negative net charge is a valid result; it indicates the net direction relative to the chosen reference.
Using the absolute value of the current would answer a different question about charge transferred without regard to direction. For net charge, retain the sign of \(I(t)\) in the integral.
A Reliable Setup
Before calculating, identify the interval, the current's units, and the reference direction. If the current is piecewise-defined, mark the interval endpoints where its formula changes. Then integrate the signed current and interpret the sign of the final value. Keep time units consistent: if \(I\) is in amperes but time is given in milliseconds, convert the time to seconds or convert the resulting charge units appropriately.
Use the start and end times for the requested period, in consistent units.
Keep positive and negative values as given; split the integral at formula changes when needed.
Compute the definite integral, report coulombs, and use the sign to state the net direction.
Differentiate an accumulated-charge function if one was used, verify the signed contributions, and confirm the units.
The central distinction is between a rate and an accumulation. Current is charge per unit time; its definite integral over time is net charge. Because direction is encoded by the sign of current, a correct calculation preserves that sign from setup through interpretation.
Check Your Understanding
Use signed current throughout, and include units in numerical answers.
- If \(I(t)=4\) amperes for \(3\) seconds, what is the net charge transferred?
- For \(I(t)=t-2\) amperes on \(0\leq t\leq 4\), calculate the net charge and identify where the current changes sign.
- A current is \(3\) amperes for the first \(2\) seconds and \(-1\) ampere for the next \(5\) seconds. What is the net charge over the full interval?
- Why does a continuous current with opposite signs at two times have to be zero at some intermediate time?
- Can a discontinuous current change sign without taking the value zero? Give a brief explanation.
