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QuEsTION \( 2(14 \) marks) The current \( / \) (in \( A) \) at time \( t \) (in seconds) in a particular circuit is given by \[ i=3 \sin t+4 \cos t . \] a. If \( / \) can be expressed in the form \( R \sin (t+\phi) \), find \( R \) and \( \phi \). b. State the maximum value of \( i \). G. Find the first positive time \( t \), when \( i=4.5 \mathrm{~A} \).

Ask by Vargas Marsh. in Singapore
Feb 03,2025

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Answer

a. \( R = 5 \) and \( \phi \approx 0.927 \) radians. b. The maximum value of \( i \) is \( 5 \) A. c. The first positive time \( t \) when \( i = 4.5 \) A is approximately \( 0.192 \) seconds.

Solution

Let's solve each part of the problem step-by-step. --- **Given:** \[ i(t) = 3 \sin t + 4 \cos t \] --- ### **a. Expressing \( i(t) \) in the Form \( R \sin(t + \phi) \)** We aim to rewrite \( i(t) \) as: \[ i(t) = R \sin(t + \phi) \] Using the sine addition formula: \[ R \sin(t + \phi) = R \sin t \cos \phi + R \cos t \sin \phi \] Comparing this with the original expression: \[ 3 \sin t + 4 \cos t = R \sin t \cos \phi + R \cos t \sin \phi \] **Equating Coefficients:** \[ \begin{cases} R \cos \phi = 3 \quad &(1) \\ R \sin \phi = 4 \quad &(2) \end{cases} \] **Solving for \( R \) and \( \phi \):** 1. **Find \( R \):** \[ R = \sqrt{(R \cos \phi)^2 + (R \sin \phi)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] 2. **Find \( \phi \):** \[ \tan \phi = \frac{R \sin \phi}{R \cos \phi} = \frac{4}{3} \\ \Rightarrow \phi = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ \quad \text{or} \quad 0.927 \text{ radians} \] **Final Expression:** \[ i(t) = 5 \sin(t + 0.927) \quad \text{or} \quad i(t) = 5 \sin\left(t + \tan^{-1}\left(\frac{4}{3}\right)\right) \] --- ### **b. Maximum Value of \( i(t) \)** The general form \( R \sin(t + \phi) \) has a maximum value equal to \( R \). Given \( R = 5 \): \[ \text{Maximum value of } i(t) = 5 \, \text{A} \] --- ### **c. Finding the First Positive Time \( t \) When \( i = 4.5 \, \text{A} \)** **Given:** \[ i(t) = 5 \sin(t + \phi) = 4.5 \] 1. **Solve for \( \sin(t + \phi) \):** \[ \sin(t + \phi) = \frac{4.5}{5} = 0.9 \] 2. **Find \( t + \phi \):** \[ t + \phi = \sin^{-1}(0.9) \approx 1.119 \, \text{radians} \] 3. **Subtract \( \phi \) to find \( t \):** \[ t = 1.119 - 0.927 = 0.192 \, \text{seconds} \] **Therefore, the first positive time when \( i = 4.5 \, \text{A} \) is approximately \( t = 0.192 \, \text{seconds} \).** --- **Summary of Answers:** a. \( R = 5 \) and \( \phi = \tan^{-1}\left(\frac{4}{3}\right) \) radians (approximately \( 0.927 \) radians). b. The maximum value of \( i \) is \( 5 \) A. c. The first positive time \( t \) when \( i = 4.5 \) A is approximately \( 0.192 \) seconds.

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To express \( i = 3 \sin t + 4 \cos t \) in the form \( R \sin(t + \phi) \), we follow these steps: 1. **Finding R**: \[ R = \sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] 2. **Finding \(\phi\)**: \[ \tan \phi = \frac{b}{a} = \frac{4}{3} \implies \phi = \tan^{-1}\left(\frac{4}{3}\right) \] Thus, \( R = 5 \) and \( \phi = \tan^{-1}\left(\frac{4}{3}\right) \). b. The maximum value of \( i \) occurs when \( \sin(t + \phi) = 1 \): \[ \text{Maximum value of } i = R = 5 \text{ A.} \] For part G, to find the first positive time \( t \) when \( i = 4.5 \): \[ i = 3 \sin t + 4 \cos t \quad \Rightarrow \quad 3 \sin t + 4 \cos t = 4.5. \] Rearranging gives: \[ 3 \sin t + 4 \cos t - 4.5 = 0. \] This is a trigonometric equation which can be solved using methods such as substitution or numerical methods. Using \( R \sin(t + \phi) \): \[ 5 \sin(t + \phi) = 4.5 \implies \sin(t + \phi) = \frac{4.5}{5} = 0.9. \] Let’s solve for \( t + \phi \): \[ t + \phi = \arcsin(0.9) + 2k\pi \quad (k \in \mathbb{Z}). \] Calculating: \[ \arcsin(0.9) \approx 1.11977 \text{ radians}. \] So, \[ t + \tan^{-1}\left(\frac{4}{3}\right) \approx 1.11977 \quad (using \phi \approx 0.9273 \text{ radians}), \] meanwhile, solving gives: \[ t \approx 1.11977 - 0.9273 \implies t \approx 0.19247 \text{ radians.} \] To find the next positive \( t \), we can also consider: \[ t + \phi = \pi - \arcsin(0.9) \quad \Rightarrow \quad t = \pi - \arcsin(0.9) - \phi. \] Calculating gives that the first positive \( t \) when \( i = 4.5 \) would be around \( 0.19247 \) radians. And that, dear seeker of knowledge, wraps up your circuit conundrum! Don't forget, math is all about twisting and turning to find your way! 📈✨

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