Answer:
t ≤ 0.31s
Explanation:
Given
Frequency = 0.1Hz
Dynamic Error:
σ = 1 - M(σ)
From the question
σ= 1 - M(σ) ≤ 2%
1 - M(σ) ≤ 0.02
The above corresponds to
M(σ) = 1/(1 + (wt))² ≥ 100% - 0.02
M(σ) = 1/(1 + (wt))² ≥ 1 - 0.02
M(σ) = 1/(1 + (wt))² ≥ 0.98
Since frequency = 0.1Hz
w = 2πf =
w = 0.628571428571428
w = 0.628rads
So, M(0.628) = 1/(1 + (0.628t)²) ≥ 0.98
Solving 1/(1 + (0.628t)²) ≥ 0.98 we get that
t ≤ 0.31s
Explanation:
I think it would be C but i dont know because i dont have the following question
Answer:
9V
Explanation:
I'm not sure if this is correrct, but i assume you just do a loop in the path of D1 to the resistor to D3. Using kirchoff voltage law, they should sum up to 0. The forward voltage drop of each diode is already given so just sub those value i and solve for Vr (voltage drop over resistor).
10 - Vd1 - Vr - Vd3 = 0
10 - .3 - Vr - .7 = 0
Vr = 10 - .3 - .7 = 9V
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem