Answer:
Explanation:
Inductance = 250 mH = 250 / 1000 = 0.25 H
capacitance = 4.40 µF = 4.4 × 10⁻⁶ F ( µ = 10⁻⁶)
ΔVmax = 240, f frequency = 50Hz and I max = 110 mA = 110 /1000 = 0.11A
a) inductive reactance = 2πfl = 2 × 3.142 × 50 × 0.25 H =78.55 ohms
b) capacitive reactance = = 1 / ( 2 × 3.142× 50 × 4.4 × 10⁻⁶ ) = 723.34 ohms
c) impedance = = 240 / 0.11 = 2181.82 ohms
Answer:
1791 secs ≈ 29.85 minutes
Explanation:
( Initial temperature of slab ) T1 = 300° C
temperature of water ( Ts ) = 25°C
T2 ( final temp of slab ) = 50°C
distance between slab and water jet = 25 mm
<u>Determine how long it will take to reach T2</u>
First calculate the thermal diffusivity
∝ = 50 / ( 7800 * 480 ) = 1.34 * 10^-5 m^2/s
<u>next express Temp as a function of time </u>
T( 25 mm , t ) = 50°C
next calculate the time required for the slab to reach 50°C at a distance of 25mm
attached below is the remaining part of the detailed solution
Answer:
attached below
Explanation:
a) G(s) = 1 / s( s+2)(s + 4 )
Bode asymptotic magnitude and asymptotic phase plots
attached below
b) G(s) = (s+5)/(s+2)(s+4)
phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4
attached below
c) G(s)= (s+3)(s+5)/s(s+2)(s+4)
solution attached below
