solut4 - i 2 in the anti-clockwise direction (the sign is...

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Some solutions for Homework 4 3.6: 12. In order to write down an ODE for this chain, consider both currents, i 1 and i 2 , indicated in Fgure 3.37. By Kirchho±’s law we need to add up all the voltages around a loop. We choose two loops: the left one and the right one. Going around the left loop, we get 2 i 1 + 5( i 0 1 - i 0 2 ) + 3 i 1 = E ( t ) . Note that there are two currents ²owing through 5 H : the current i 1 in the clockwise direction (thus the sign is positive), and
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Unformatted text preview: i 2 in the anti-clockwise direction (the sign is negative). The equation for the other loop is this: 10 i 2 + 4 i 2 + 5( i 2-i 1 ) = 0 . It is zero because there is no source in that loop. Again, through 5 H we have the current i 2 in the clockwise direction and i 1 in the anti-clockwise direction. Once you have this system, you can solve it in the usual way by taking the Laplace transform. 1...
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This note was uploaded on 01/12/2010 for the course MATH 421 taught by Professor Nataliakomarova during the Winter '07 term at San Jose State University .

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