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Kobotan [32]
3 years ago
6

For the given network, if R = 0.8 [Ω], the natural response of i in(t) is:

Engineering
1 answer:
11111nata11111 [884]3 years ago
6 0

Answer:

Iin(t) =1.3 × (-0.8/1.56) e^-t/1.56 A

Explanation:

In physics, the determination of the term " natural response" simply means that we want to know what happens in a circuit when the value of t = 0, that is to say after the circuit has been disconnected. Hence, the value of the voltage and the current can then be determined or Calculated;

For the the natural response of i in(t) we will be using the formula below;

I(t) = Vo × t/ R = Vo/R × e^-t/h.

Where h = 1/RC = time constant.

For t= 0^- = 0.8 × 1= 0.8 V.

1/Ctotal = 1/ 2 + 1/3 = 6/5

For t = 0^+;

h =( 0.8 + 0.5) × 6/5 = 1.56 seconds.

Hence, we will have;

Vin(t) = 0.8 × e^-t/1.56.

Iin(t) =1.3 × (-0.8/1.56) e^-t/1.56

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Answer:

a)  ∝  and β

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b)

The phase is; ∝  

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Explanation:

a) 15 wt% Sn - 85 wt% Pb at 100⁰C.

The phases are ; ∝  and β

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C_{\beta } =  98 wt% Sn - 2wt% Pb

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Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%

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Answer:

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Answer:

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