The switch in the circuit shown below has been in the position \'a\' for a long
ID: 1798997 • Letter: T
Question
The switch in the circuit shown below has been in the position 'a' for a long time. In other words, the circuit has reached a steady-state condition. Then at t=0, the switch instantaneously moves to position 'b'. If it is given that Ro=7 ? and Lo=41 mH, then what is the voltage vo(t) as t=0.5 ms ? Express your answer in Volts to three decimal places.You should use a MATLAB/Octave script to calculate your answer to avoid simple errors in calculation. Note that in MATLAB e-tis given by exp(-t).
The switch in the circuit shown below has been in the position 'a' for a long time. In other words, the circuit has reached a steady-state condition. Then at t=0, the switch instantaneously moves to position 'b'. If it is given that Ro=7 ? and Lo=41 mH, then what is the voltage vo(t) as t=0.5 ms ? Express your answer in Volts to three decimal places.You should use a MATLAB/Octave script to calculate your answer to avoid simple errors in calculation. Note that in MATLAB e-tis given by exp(-t).Explanation / Answer
consider when the switch is in position 'b'
let the two loop currents be i1 and i2
apply KVL in the loop 1:
800=120i1+40(i1-i2) --------------------(1)
apply KVl in loopp 2:
40(i2-i1)+7i2+(41x10^(-3))di2/dt=0 -----(2)
from eq(1):
i1 = 5 + (i2/4)
substitute in eq(2) and solve it:
di2/dt + 902i2 = 4878
apply laplace transform to it:
[sI2(s)-I(0)] + 902I2(s) = 4878/s ---------(3)
I(0) can be found out when the switch is in position 'a'
apply current division rule and you can have
I(0) = 4.26
substitute I(0) in eq(3) and solve it:
I2(s) = (4878+4.26s)/[s(s+902)]
apply partial fractions and solve it:
I2(s) = 5.41/s - 1.15/(s+902)
apply inverse laplace:
i2 = 5.41-1.15e^(-902t)
i1= 5 + (i2/4) = 6.35 - 0.29e^(-902t)
now the current passing through 40ohms after changing the switch position is
I = i1-i2 = 0.9+0.86e^(-902t)
now the voltage Vo is given by
Vo(t) = 40I = 36+34.4e^(-902t)
Vo(0.5ms) = 36+34.4e^(-902x0.5x10^(3)) = 57.91V
I guess there is slight change in the decimal points.. please ask me if you've any doubt... I'm here to help you and myself..
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