The currents in the electrical network in Figure 6.1
Chapter 6, Problem 40E(choose chapter or problem)
The currents in the electrical network in Figure satisfy the system
\(\frac{1}{9} I_{1}+64 I_{2}^{\prime \prime}=-2 \sin \frac{t}{24} \frac{1}{64} I_{3}+9 I_{3}^{\prime \prime}-64 I_{2}^{\prime \prime}=0 I_{1}=I_{2}+I_{3}\)
Where \(I_{1}, I_{2}\), and \(I_{3}\) are the currents through the different branches of the network. Using the elimination method of Section , determine the currents if initially \(I_{1}(0)=I_{2}(0)=I_{3}(0)=0, I_{1}(0)=73 / 12, I_{2}(0)=3 / 4\), and \(I_{3}^{\prime}(0)=16 / 3\)
Figure 6.1 An electrical network
Equation Transcription:
Text Transcription:
1/9 I_1+6/4 I_2^′′ =−2sin t/24 1/64 I_3+9I_3^′′−64 I_2^′′ =0 I_1 =I_2+I_3
I_1
I_2
I_3
I_1 (0)=I_2 (0)=I_3 (0)=0,I_1 (0)=73/12, I_2^′ (0)=3/4
I_3^′(0)=16/3
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