Problem 23 - Basic Technical Mathematics with Calculus

Question

In an amplifier circuit, the current \(i\) (in A) changes with time \(t\) (in s) according to \(i=0.06 t \sqrt{1+t^{2}} .\) If \(0.015 \mathrm{C}\) of charge has passed a point in the circuit at \(t=0,\) find the total charge to have passed the point at \(t=0.25 \mathrm{s}\)

Step-by-Step Solution

Answer

The total charge at \(t = 0.25\) s is approximately 0.0171 C.

Step 1: Determine the Integral for Charge

The current passing through a point in the circuit is given as a function of time: \(i(t) = 0.06 t \sqrt{1+t^2}\). To find the total charge that has passed by time \(t\), we need to integrate the current function \(i(t)\) with respect to time. The amount of charge \(Q\) at time \(t\) is given by the integral: \[ Q(t) = \int i(t) \, dt = \int 0.06 t \sqrt{1 + t^2} \, dt \] Next, we need to evaluate this integral from the initial time to \(t = 0.25s\).

Step 2: Evaluate the Integral

To solve the integral \(\int 0.06 t \sqrt{1 + t^2} \, dt\), it can be helpful to use a substitution method. Let \( u = 1 + t^2 \), then \( du = 2t \, dt \). Solving for \(t \, dt\), we get: \[ t \, dt = \frac{1}{2} du \].This changes the integral into: \[ \int 0.06 \times \frac{1}{2} \sqrt{u} \, du = 0.03 \int \sqrt{u} \, du \]Evaluating the integral \(0.03 \int u^{1/2} \, du \) results in: \[ 0.03 \times \frac{2}{3} u^{3/2} = 0.02 u^{3/2} \]Substituting back \( u = 1 + t^2 \), we have:\( 0.02 (1 + t^2)^{3/2} \).

Step 3: Integrate with Limits

Now we will evaluate the integral from the lower limit \(t = 0\) to the upper limit \(t = 0.25\) seconds.\[ Q(0.25) = 0.02 \left[ (1 + (0.25)^2)^{3/2} - (1 + 0^2)^{3/2} \right] \]Calculate each term:- For \(t=0.25\): \((1 + (0.25)^2)^{3/2} = (1 + 0.0625)^{3/2} = (1.0625)^{3/2}\)Use a calculator to find this value.- For \(t = 0\), it simplifies to 1.So the integral evaluates to:\[ Q(0.25) = 0.02 ((1.0625)^{3/2} - 1) \]

Step 4: Compute the Numeric Result

Now compute \((1.0625)^{3/2}\), which approximately equals 1.1035.Substitute back into the equation:\[ Q(0.25) = 0.02 (1.1035 - 1) = 0.02 \times 0.1035 = 0.00207 \text{ C} \]Add the initial charge of 0.015 C given at \(t = 0\): \[ Q_{total} = 0.015 + 0.00207 = 0.01707 \text{ C} \]

Step 5: Conclusion

Therefore, the total charge that has passed the point at \(t = 0.25 \text{s}\) is approximately \(0.0171 \text{ C}\).

Key Concepts

Substitution Method

The substitution method is a powerful technique in integral calculus, used to simplify the process of finding integrals. When dealing with complex functions, substitution can make otherwise challenging integrals more manageable.

In this particular exercise, our goal is to integrate the current function over time. The given current function is:

• \(i(t) = 0.06 t \sqrt{1+t^2} \)

Using substitution, we aim to simplify this equation to perform the integration more easily.

How It Works

To substitute, we find a part of the function that, when replaced with a simpler variable, will make integration straightforward. We set:

• \( u = 1 + t^2 \)
• Then, \( du = 2t \, dt \)

By rearranging, \( t \, dt \) becomes \( \frac{1}{2} \, du \), simplifying our integral to involve only \( u \),

• \( 0.06 \times \frac{1}{2} \sqrt{u} \, du = 0.03 \int \sqrt{u} \, du \)

This method transforms the problem into a simpler one, which we can solve to find \[ 0.02 u^{3/2} \], allowing us to focus on solving the integral rather than the initial complex function.

Current and Charge Relationship

In electrical circuits, the relationship between current and charge is fundamental. Current \( i(t) \) is the rate at which charge \( Q \) flows through a point in the circuit. This relationship can be expressed in a key equation:

• \( Q(t) = \int i(t) \, dt \)

This means that to find how much charge has moved past a point over time, we can integrate the current function.

Understanding Current Function

The current provided in the problem is \(i(t) = 0.06 t \sqrt{1+t^2}\), showing that the current changes with time. Integrating this function from a starting point to a specific time will give us the total charge that has flowed up to that time.

For example, integrating this function from \(t=0\) to \(t=0.25s\) computes the net amount of charge passing through the circuit within that timeframe. This transformation from a rate (current) to a cumulative quantity (charge) highlights the importance of integration in understanding electric circuits.

Integration with Limits

Integration with limits involves evaluating an integral over a specific interval, providing meaningful real-world implications particularly in physics and engineering.

After simplifying the integral using substitution, the next step was integration with limits from \(t = 0\) to \(t = 0.25s\). We find:

• \( Q(0.25) = 0.02 \left[ (1.0625)^{3/2} - 1 \right] \)

This specific integration calculates the total charge transferred within the circuit during the specified time interval.

Numerical Evaluation

By tracking how the current function evolves over time and integrating it within these precise limits, we assess the electric charge passing through the point in the circuit. Substituting specific values for \(t\), we get the precise amount of charge that passed into the circuit between the initial and final times; this calculation yields approximately 0.00207 C.Adding this result to the initial 0.015 C existing at \(t=0\), we conclude with a total charge of 0.0171 C by time \(t=0.25\) seconds.