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Chapter 1: Problem 13

The current through a particular circuit element is given by \(i(t)=10 \sin (200 m t) A\) in which \(t\) is in seconds and the angle is in radians. a. Sketch \(i(t)\) to scale versus time for \(t\) ranging from 0 to \(15 \mathrm{ms}\) b. Determine the net charge that passes through the element between \(t=0\) and \(t=10 \mathrm{ms}\), c. Repeat for the interval from \(t=0\) to \(t=15 \mathrm{ms}\).

Short Answer

Expert verified
Sketch is sinusoidal; net charges are 0 C (0-10ms) and 0.0318 C (0-15ms).

Step by step solution

01

Understanding the Problem Statement

The function given is the current, \(i(t) = 10 \sin(200 \pi t)\) A. We need to sketch this function, determine the charge between given time intervals of \(0\) to \(10 \mathrm{ms}\) and \(0\) to \(15 \mathrm{ms}\).
02

Sketching the Current over Time

To sketch \(i(t) = 10 \sin(200 \pi t)\), note that it's a sinusoidal function with an amplitude of 10 A. The period \(T\) can be calculated using \(T = \frac{1}{f}\) where \(f\) is the frequency. Here, \(f = \frac{200 \pi}{2 \pi} = 100\) Hz, so \(T = 0.01\, \mathrm{s} = 10\, \mathrm{ms}\). This function has maximum values at \(10\, \mathrm{A}\) and minimum values at \(-10\, \mathrm{A}\) alternating sinusoidally every \(10\, \mathrm{ms}\). Sketch this waveform from \(t = 0\, \mathrm{ms}\) to \(t = 15\, \mathrm{ms}\), showing one and a half periods.
03

Calculating Charge from 0 to 10ms

The charge \(Q\) flowing through a circuit element is given by the integral of the current over time, \(Q = \int_{0}^{T} i(t) \, dt\). For \(t = 0\) to \(t = 10 \mathrm{ms}\), \(Q = \int_{0}^{0.01} 10 \sin(200 \pi t) \, dt\). This evaluates to:\[Q = \left[-\frac{10}{200 \pi} \cos(200 \pi t)\right]_{0}^{0.01} = \left[-\frac{10}{200 \pi} (\cos(200 \pi \times 0.01) - \cos(0))\right]\]\[= \left[-\frac{10}{200 \pi} (\cos(2 \pi) - 1)\right] = \left[-\frac{10}{200 \pi} (1 - 1)\right] = 0 \; \mathrm{C}\].
04

Calculating Charge from 0 to 15ms

For \(t = 0\) to \(t = 15 \mathrm{ms}\), we integrate from \(0 \) to \(0.015 \) seconds. So,\[Q = \int_{0}^{0.015} 10 \sin(200 \pi t) \, dt\]\[Q = \left[-\frac{10}{200 \pi} \cos(200 \pi t)\right]_{0}^{0.015} = \left[-\frac{10}{200 \pi} (\cos(3 \pi) - 1)\right]\]\[= \left[-\frac{10}{200 \pi} (-1 - 1)\right] = \frac{20}{200 \pi} = \frac{1}{10 \pi} \approx 0.0318\, \mathrm{C} \].
05

Conclusion

The sketch represents a sinusoidal wave from 0 to 15 ms. From 0 to 10 ms, the net charge is 0 C, and from 0 to 15 ms, the net charge is approximately 0.0318 C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Waveform Sketching
When sketching the waveform of a sinusoidal function like \(i(t) = 10 \sin(200 \pi t)\), it helps to understand its basic characteristics:
  • Amplitude: The amplitude of the sinusoidal function here is 10 A. This means the wave oscillates between +10 A and -10 A.
  • Period: The period \(T\) is the time it takes for the wave to complete one full cycle. It is calculated as \( T = \frac{1}{f} \), where \( f \) is the frequency. For this function, the frequency is determined by \( \frac{200 \pi}{2 \pi} = 100 \) Hz, so the period \(T\) is 0.01 seconds or 10 milliseconds.
  • Sketching: You start at 0 ms with \(i(t) = 0\), climb to 10 A at 2.5 ms, return to 0 A at 5 ms, fall to -10 A at 7.5 ms, and return to 0 at 10 ms, completing one full cycle. From 10 ms to 15 ms, it repeats starting again from 0 A, representing half a cycle.
Using these points, you can plot the sinusoidal function, showing clear peaks, troughs, and zero crossings at the appropriate intervals.
Charge Calculation
To determine the net charge that flows through the circuit between specific time intervals, we use the integral of the current over time. Here's how it plays out:
  • Charge Formula: The charge \(Q\) is defined by the integral \( Q = \int_{t_1}^{t_2} i(t) \, dt \). This represents the area under the current-time graph between two times \(t_1\) and \(t_2\).
  • Calculation from 0 to 10ms: For time \(t = 0\) to \(t = 10 \mathrm{ms}\), the wave completes exactly one period. Since the sinusoidal wave is symmetrical, the areas above and below the time axis are equal, canceling each other out, resulting in a net charge of 0 C.
  • Calculation from 0 to 15ms: For \(t = 0\) to \(t = 15 \mathrm{ms}\), one and a half cycles are considered. Over the additional half cycle (from 10ms to 15ms), there is a net positive area under the curve, resulting in a calculated charge of approximately 0.0318 C.
Thus, careful calculation over each segment of interest provides clear insights into the charge transfer over time.
Integral of Sinusoidal Function
Understanding the integral of a sinusoidal function like \(10 \sin(200 \pi t)\) is crucial for calculating charge. Here鈥檚 how we handle it:
  • Integration Process: The integral \( \int 10 \sin(200 \pi t) \, dt \) yields \(-\frac{10}{200 \pi} \cos(200 \pi t) + C\), where \(C\) is the constant of integration. It turns the sinusoidal function into one involving cosine.
  • Boundary Evaluation: For a fixed interval, evaluate the antiderivative at the upper and lower bounds and subtract: \[ Q = \left[-\frac{10}{200 \pi} \cos(200 \pi \times t_2) + C\right] - \left[-\frac{10}{200 \pi} \cos(200 \pi \times t_1) + C\right] \].
  • Specific Intervals: Applying this to both 0 to 10 ms and 0 to 15 ms gives precise values reflecting the net charge 鈥� 0 C and 0.0318 C, respectively, highlighting symmetry and area concepts when interpreting sinusoidal functions over their cycles.
Integrating sinusoidal functions can yield valuable information about electrical circuits in a variety of time intervals.

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Most popular questions from this chapter

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