Electric Circuits: Worked Example Walkthroughs

Abstract

This article develops the relationship between charge and current in electric circuits through calculus, then demonstrates how to locate maximum current in transient responses. We work through the integral relationship connecting these quantities and apply optimization techniques to find peak current values—skills essential for circuit design and component selection.

Background

In circuit analysis, two fundamental quantities are current and charge. Current $i(t)$ represents the instantaneous rate at which charge flows through a conductor, measured in amperes (coulombs per second). Charge $q(t)$ represents the total amount of electric charge that has accumulated or flowed, measured in coulombs.

These quantities are related through calculus: current is the time derivative of charge, and conversely, charge is the time integral of current. This relationship underpins our ability to predict circuit behavior, especially in components like capacitors that store charge and respond to accumulated electrical energy.

Understanding how to move between current and charge representations, and how to find critical points like maximum current, is essential for practical circuit design. Engineers must know peak currents to select wire gauges, fuses, and component ratings that prevent damage during transient events.

Key Results

Charge from Current via Integration

The total charge $q(t)$ that has flowed through a circuit element up to time $t$ is obtained by integrating the current function ^{[charge-as-a-function-of-current]}:

$q(t) = \int_0^t i(x) \, dx$

This integral "sums" the infinitesimal charge contributions $i(x) \, dx$ across the time interval from $0$ to $t$. The relationship reflects the fundamental definition of current as charge per unit time: reversing differentiation via integration recovers the accumulated charge.

Finding Maximum Current

For circuits exhibiting exponential transient behavior, the current function typically has a peak value that occurs at a specific time. To find this maximum, we differentiate the charge function to obtain current, then set $\frac{di}{dt} = 0$ and solve for time ^{[maximum-current-in-a-circuit]}.

For charge functions characterized by a time constant or decay parameter $\alpha$, the maximum current occurs at:

$t_{max} = \frac{1}{\alpha}$

with the maximum current value:

$i_{max} = \frac{1}{\alpha} e^{-1}$

This result is typical of first-order RC or RL circuits, where current exhibits transient overshoot before settling to steady state.

Worked Examples

Example 1: Charge Accumulation from a Time-Varying Current

Problem: A circuit element carries a current $i(t) = 5e^{-2t}$ amperes for $t \geq 0$. Find the total charge that has flowed through the element by time $t = 1$ second.

Solution:

Using the integral relationship ^{[charge-as-a-function-of-current]}, we compute:

$q(1) = \int_0^1 5e^{-2x} \, dx$

Evaluate the antiderivative:

$q(1) = 5 \left[ -\frac{1}{2}e^{-2x} \right]_0^1 = -\frac{5}{2} \left( e^{-2} - 1 \right)$

$q(1) = \frac{5}{2} \left( 1 - e^{-2} \right) = \frac{5}{2}(1 - 0.1353) \approx 2.16 \text{ coulombs}$

Interpretation: By $t = 1$ second, approximately 2.16 coulombs have flowed through the element. The exponential decay of current means that most charge flows early; as time increases further, the additional charge accumulation slows.

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Example 2: Finding Maximum Current in a Transient Response

Problem: In an RC circuit, the charge on the capacitor evolves as $q(t) = Q_0(1 - e^{-t/\tau})$ where $Q_0 = 10$ coulombs and $\tau = 0.5$ seconds. Find the time at which current is maximum and compute that maximum value.

Solution:

First, obtain the current by differentiating the charge:

$i(t) = \frac{dq}{dt} = \frac{d}{dt}\left[Q_0(1 - e^{-t/\tau})\right] = Q_0 \cdot \frac{1}{\tau} e^{-t/\tau}$

$i(t) = \frac{10}{0.5} e^{-t/0.5} = 20 e^{-2t}$

Here, $\alpha = 2$ (the decay rate). To find the maximum, differentiate current:

$\frac{di}{dt} = 20 \cdot (-2) e^{-2t} = -40 e^{-2t}$

Since $e^{-2t} > 0$ for all $t$, we have $\frac{di}{dt} < 0$ for all $t \geq 0$. This means current is monotonically decreasing—the maximum occurs at $t = 0$.

$i_{max} = i(0) = 20 \text{ amperes}$

Interpretation: This result is typical for a capacitor charging from an initial condition. The current is largest at $t = 0$ when the voltage across the capacitor is zero, and decays exponentially as the capacitor charges. There is no interior maximum; the peak current is the initial current.

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Example 3: Maximum Current with a Different Charge Profile

Problem: Consider a circuit where charge accumulates as $q(t) = At e^{-\alpha t}$ with $A = 5$ coulombs/second and $\alpha = 1$ s$^{-1}$. Find the time and value of maximum current.

Solution:

Differentiate to obtain current:

$i(t) = \frac{dq}{dt} = A e^{-\alpha t} + At \cdot (-\alpha) e^{-\alpha t} = A e^{-\alpha t}(1 - \alpha t)$

$i(t) = 5 e^{-t}(1 - t)$

To find the maximum, set $\frac{di}{dt} = 0$:

$\frac{di}{dt} = 5 \left[ -e^{-t}(1 - t) + e^{-t}(-1) \right] = 5 e^{-t} \left[ -(1 - t) - 1 \right] = 5 e^{-t}(t - 2)$

Setting this equal to zero: $t - 2 = 0$, so $t_{max} = 2$ seconds.

The maximum current is:

$i_{max} = 5 e^{-2}(1 - 2) = -5 e^{-2} \approx -0.677 \text{ amperes}$

Interpretation: The negative sign indicates that current flows in the reverse direction at the maximum. This behavior arises in circuits with competing charging and discharging mechanisms. The peak magnitude occurs at $t = 2$ seconds, and component ratings must accommodate this reverse current to prevent damage.

References

• ^{[charge-as-a-function-of-current]}
• ^{[maximum-current-in-a-circuit]}

AI Disclosure

This article was drafted with the assistance of an AI language model. The mathematical derivations, worked examples, and explanations were generated and organized by the AI based on the provided course notes. All factual claims and mathematical statements are cited to the original note sources. The author reviewed the content for technical accuracy and clarity before publication.